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CONTINUITY 



IN 



CONCRETE 
BUILDINQ FRflflES 



Practical Analysis for 
Vertical Load and Wind Pressure 



PORTLAND CEMENT ASSOCIATION 





' 






CONTINUITY 



IN 



CONCRETE 
BUILDINQ FRdnES 



;• 









Tribune Building 




Montevideo, S* A* 



Published by 

PORTLAND CEMENT ASSOCIATION 



Concrete for Permanenc 



e 



33 WEST GRAND AVENUE . CHICAGO, ILLINOIS 



[BLANK PAGE] 




CCA 



INWH NATIQNAJ 






PREFACE 

The common practice of designing beams in building frames by use of 
moment coefficients and ignoring moments in columns is not always safe 
and adequate. It may result in distress in many structures and over-design 
in others. It disregards many important elements including concentrated 
loads, ratio of live load to dead load, ratio of span lengths in adjacent 
bays, ratio of column stiffness to beam stiffness, and ratio of column width 
to span length. The effect of all these elements upon design cannot be incor- 
porated in three simple beam moments such as l /i Q wl 2 , l /nwl 2 and l /uwl 2 . 

One reason why the moment coefficients are still used is that making a 
proper analysis has been a complex and lengthy task. This booklet is 
prepared 10 facilitate the analysis and to discuss quite fully the elements 
that go into the making of a well-balanced design. More specifically, the aim 
is to give constructive suggestions for improved practice and to present 
analytical procedures that are practical, i.e., quick, easy, convenient and 
yet sufficiently accurate for office design practice. 

Considering the reduction in labor by the use of the proposed procedures, 
which are approximate, the accuracy of the results is surprisingly good. 
Even procedures that are mathematically exact do not yield exact results 
because they rest on assumptions that are approximate. In fact, better 
accuracy may be obtained from "approximate" analysis based on good 
assumptions than from "exact" analysis based on poor assumptions. 

Good judgment and thorough understanding of the assumptions under- 
lying the analysis are essential For this reason, the presentation goes 
beyond the mere statement of analytical procedures, which are simple and 
occupy limited space only. Included are also: underlying assumptions, dis- 
cussion and special studies, most of the latter in form of numerical prob- 
lems. Deriv ii ions that are not essential to the application of the procedures 
have been deferred to the appendixes. 

The problems are recommended for careful study because they make 
the render familiar with the simplicity of the procedure, the accuracy 
obtainable and the effect of varying assumptions. 

Procedures are given for both vertical load and wind pressure. The dis- 
cussion of wind pressure applies to earthquake analysis also, if earthquake 
shocks are treated in the customary manner as static horizontal loading. 

The analysis presented for vertical load is essentially Professor Cross' 
method of moment distribution applied— with slight modifications — to two 
joints only. For very irregular cases, such as offset columns supported on 
g transfer girders, the procedure may have to be extended by using moment 

distribution applied to more than two joints. Also for wind pressure, the 
highly irregular cases need special study beyond that given in this text. 



3 



O 
P 



December. 1935 



L\ 



[BLANK PAGE] 




CCA 



IN T F NNATtONAl 






CONTENTS 



NOTATION 
VERTICAL LOAD 

General Procedure of Analysis 

1. Introduction ......... 

2. General procedure of analysis . 

3. Selection of Q-values when cross-sections are not known 

4. Examples of analysis ........ 

5. Moments in beams in roofs and bottom floors 

6. Shear in beams ........ 

7. Column moments and beam moments at wall columns 

Special Procedure of Analysis 

8. Assumptions regarding restraint at far ends of columns 

9. Correction for rotation at far ends of columns 

10. Summary ) 

Discussion of Various Effects 

11. Effect of width of columns ...... 

12. Effect of change in moment of inertia .... 

13. Effect of distant loads ....... 

It. Effect of haunching ........ 



Page 

5 

6 

7 

9 
15 
15 
16 



19 
20 

23 



24 
26 
29 
29 



WIND PRESSURE 

15. Introduction ...... # ... 31 

16. Illustrative problem ........ 32 

17. Eccentric wind pressure ....... 35 

18. Accuracy of approximate analysis ...... 37 

19. Moment deflection and the ideal layout .... 39 

20. Illustrative problem^ vertical load and wind pressure combined 40 

APPENDIX A: Derivation of Formulas— Vertical Load 

21. Angle changes determined by moment area principle . . 45 

22. Angle changes due to end moments ..... 46 

23. End moments required for fixity ...... 47 

24. Moments as a function of end rotation and fixed end moments 48 

25. Joint rotation a function of stiffness ..... 49 

26. Derivation of formulas for joint coefficient, Q . . . 50 

27. Conventions and use of signs ...... 51 

28. Formulas for moments in columns ..... 53 

APPENDIX B: Derivation of Formulas— Wind Pressure 

29. Assumptions and distribution of wind pressure ... 53 



BIBLIOGRAPHY 



NOTATION 



M = moment. 

Mab = end moment at joint A of member AB. 

M F = fixed end moment. 

U = unbalanced moment. 

W = total load on span. 

w = load per linear foot. 

P = concentrated load or axial load. 

I — span length. 

h = story height. 

J = moment of inertia. 

K = I /I or I/h. 

n = ratio of U K for column" to "K for beam" 

Q = a joint coefficient. 

e = eccentricity. 

d = depth of section. 

a = width of supporting member. 

b = width of section. 

A = area. 

c = diameter of circle. 

V = end shear. 

v = relative shear. 

= angle of rotation at end of member. 

E = modulus of elasticity. 

/ = a coefficient expressing degree of restraint 

R = angle of joint translation. 



Continuity 



in 



Concrete 
Building Frames 



Vertical Load 

General Procedure of Analysis 
1. Introduction 

A procedure is presented for determination of end moments in horizontal 
members of continuous building frames. It has certain characteristics in 
common with current analytical methods, and the underlying principles 
in the derivation — given in Appendix A — are the same as in, for example, 
the slope-deflection and the moment-distribution methods. Current ana- 
lytical methods are not superseded but merely adapted so that results with 
satisfactory degree of accuracy may be obtained by the least amount of work. 

In analysis of continuous frames, the concept of fixed end moments is 
especially useful; they are the moments required to fix the ends of members, 
or — as is sometimes said — to lock the joints in frames. Fixed end moments, 
M F , may be expressed in terms of the loading, W, and the span length, I; 
and M F for one member is independent of other members. Moment coeffi- 
cients for prismatic beams* with fixed ends and subject to nineteen types 
of loading are compiled in Fig. 1, in which the example in the lower right 
corner illustrates the use of the moment coefficients. 

While determining values of M F as, for example, in Fig. 2, joints C, 
A, B and D are assumed to be temporarily locked (prevented from rotat- 
ing). If the artificial restraint is removed, each joint will rotate under the 
influence of the unbalanced moment: the difference between the fixed end 
moments. The use of unbalanced moments — denoted as U — is convenient 
in frame analysis, and U may readily be determined from values of M F . 

When joints rotate, new moments are induced at the ends of the mem- 
bers, and the final moment is equal to the sum of M F and "induced 
moments." The problem to be solved by analysis of typical building frames 
is principally to determine the induced moments. 

The induced moments are usually small compared with M F . They may — 
in a sense — be considered as a correction to M F } and a high degree of 
accuracy in the determination of induced moments is not essential. These 
moments will be approximated in the procedure which follows, and the 
work involved in the frame analysis will thereby be greatly reduced. 

♦Members will be assumed prismatic except in Section 14: Effect of Haunching. 

5 



Coefficients of W*l 



Symmetrical loading 

i y i 



i- 



2: 



3- 



4: 



5- 



e- 



7- 



8- 



9: 



ZO- 



lh 



t 



(Jnsymmetrical loading 

W--{wl 



/ 
3 



12'- k 



10 _ 




J_ 

IS 



I 



w 




7 W 7 

3^T3 



2 
9 



f3 '9& 



i 



. 






— 






4 "j*V ~JT4 ~^T4 



Each lodd:W 



3 _±l±f±fZZLj a I 



5 

10 



a la 



I* 



a(G-8a+3a z ) 

12 



h. 



Ji 



z 

W-wal 



96 



1-3 1 



a H4-3a) 

n 



I 



- 12 



15: afl-a) z 



a! 




l-al 



aH/-a) 



W=wl 



t 



m^ 



i 



r 



ii 



16 



23 



740 






mi iff i 




,n. a(IO-IOa+3a l ) 
U ' 30 




7 



al 



l-al 



a 2 (S-3a) 



l*jk 



- 1 IV--ju/l 

w * 



l 9: 3fl0-/5a*6a*) LrtiMK 




A 

J 40 



W -Jural 



3 1 



l-al 



9H5-43) 
16 



fxamp/e : 



4rr 



•5/ 






JV-- 2wal 



T 



20 0tf<? 




IE 



I y*t 



-^4- /5 : 



± 



T 
3 



/iW7 



i 



■l-2al^l-al 



W--w{l-2all 



n 



■alA—l-Zal^ral 



a (3-7 a) 

Aa?oo*?o)?o* 4o,ooo 

k (i*800* 20)70 --16, 000 
I* 7a -2a 2 if JM20.000) 2 46,000 



Fixed End Moment 

at & 



,1(1200*20)20-40,000 

,i (1*800*20)70-- 10,100 

(iM(mOO0) 70 * 18,700 



12 



Totah 117,000 ft lb. Tot ah 69. 400 ft. lb. 



Fig. 1* 



Fixed End Moments — in Terms of Wl — for Prismatic Beams 



2. General Procedure of Analysis 

The value of M F for prismatic members is a function of loads, W, and 
span lengths, I; but an induced moment depends also upon the moment 
of inertia, /, of the members in the frame. Let K denote the ratio of I /I 
for individual members, and n the ratio of K- value of a column to if-value 



*By permission of the authors, Fig. 1 is taken — with slight modifications — from Fig. 5, page 85, in 
'Continuous Frames of Reinforced Concrete," by Cross and Morgan, published by John Wiley and 
Sons. 1932. 



6 



of a beam. Equations for n A and n B are given in Fig. 2. This figure illustrates 
the general procedure which is described in Steps (1) to (6): 

(1) Record in Fig. 2 the loading on the three adjacent spans, 1 C a, Iab 
and l BD . 

(2) From this loading, determine values of M F (using Fig. 1 ) and record 
M F as negative quantities in Fig. 2. 

(3) Compute the unbalanced moment at A, Ua, and at B, U B , as M F 
to the left minus M F to the right of each joint. Record U with 
proper sign. 

(4) Compute (or select, as explained in Section 3) values of n or Q 
(see Fig. 2 for formula for Q* applying to the case shown). 

(5) Record products of U times Q as shown and with proper signs as 
in Fig. 2. 

(6) Determ in e the final end moment as the algebraic sum of the three 
terms in the spaces marked X. 

The calculations in Steps (1), (2) and (3) are those that precede most 
methods of analysis. Steps (4), (5) and (6) comprise the analytical work 
characteristic of the procedure in this text and will be discussed further 
in subsequent sections. 



T 




I 



CA 



4- 



l AB» h 




I 



A3 






i 



so 




c 



< 



Wc-MIb 




Record the loading here 




* A 



A 



m 



*2Q**U* 



Ti 



AB 



*0m'U, 



B 



M 



M 



Q**u A 



-2Q 6 'U a 



1 




d£ 

Q = „ ' n CSee F/G.3) 



M 



£A 




D 




ML 



M 8A~ M 6D^'d 



4n*-2 



I 



8 




Negative moments give tension in top of beams. 
I A , I 3 and h are average raiues for columns above and below joints A and 5. 
Note that LI • moment to the left of joint minus moment to the right of joint. 

Fig. 2 



5. Selection of Q-Values When Cross- Sections Are Not 
Known 

The relative values of K — denoted as n — must be chosen before a frame 
can be analyzed. How to select n, and the joint coefficient Q } is a question 
of the greatest importance to the designer. 

In reviewing a frame already designed, the procedure is simply to com- 
pute K, n, Q and then proceed with the analysis. This is the type of problem 

*See Appendix A for derivation. 



usually treated in papers and textbooks. When only loads, spans and story 
heights are known, the values of Q must be selected. 

The reader who is familiar with Professor Cross' moment distribution 
method* will recognize in the procedure in Fig. 2 the same characteristics 
as in that method: fixed end moments {M F ) t unbalanced moments (U), 
carry-over factors, and distribution factors (Q). A number of cases were 

1 



studied in most of which a Q-factor of 



4n + 2 



gave better results for frames 



with two joints than the corresponding factor, , used for regular 



1 



4n+ 2 



0.5 



0.4 



0.3 



moment distribution in frames with many joints. The equations Q = 

1 

proposed for joints in floors and Q —- — ^r~z proposed in Section 5 for joints 

in roofs are derived empirically consistent with the expressions for dis- 
tribution factor known from the moment distribution method. A refine- 
ment of giving different Q-formulas for interior and exterior joints is 
deemed unwarranted. 

According to the equation in 
Fig. 2, Q equals Yi when n = (no 
column restraint) and decreases to 
zero with increasing column stiff- 
ness. The variation of Q between 
the limits of Y and zero is plotted 
in Fig. 3. 

When n is greater than about 4, 

the Q- values are small and vary 
only slightly. In this range, which 
includes the lower stories of tall 
buildings, Q = V20 is a good average 
value . 

Part of the Q-diagram near the 
ordinate axis (n = in Fig. 3) corre- 
sponds to slightly restrained con- 
tinuous girders. For ordinary building frames, n need seldom be taken 
smaller than one-half,** and the value of Q = YL may be selected as 
typical for frames with slender columns having n-values smaller than unit> 

When n is between 1 and 4, the value of Q = }/$ may be considered typ- 
ical. This is the range between the very slender and the very stubby columns. 

If cross-sections arc not known in ordinary building frames, the following 
suggestions may be used as a guide in selecting values of Q: 



0.2 



0.1- 

























( No CO) 


'umn 


' res 


fraint 










1 




















1 

\ 








Diagran 


7 IS 


V~4k 


1 
1+2 




\ 




























































5»_ 




















% 




















<^* 


n- 


valu 


es 



















4 

Fig. 3 



£ 



to 



Values of n 


Below 1 


Between 1 and 4 


Above 4 


Column Classification 


Slender 


Medium 


Stubby 


Values of Q 


M 


H 


Vm 





ee References 10 and IK 
**See CoDclusionB in Section 4. 



<'! 



4* Examples of Analy 



Problem 1. The frame in Fig. 4(a) has equal spans 20 ft. long and equal 
story heights; dead load equals 1000 lb. and live load 1200 lb. per lin. ft. 
Determine the maximum center moments in span AB and the corresponding 
end moment, M A b- 

The analysis will be carried out according to steps (1) to (6) on page 7 
and arranged as in Fig. 2. Maximum center moment in AB is produced 
by placing live load in AB and also in alternate spans as in Fig. 4(a).* 
This determines the loadings and the fixed end moments, M F . The beam 
and column dimensions are not known. Suppose the columns will be very 
slender and select Q = % (see Section 3). 

From the loading and span given, compute the fixed end moments at 
A and B by using the coefficient in case 5 in Fig. 1, which gives 

M F = - l / 12 W X I = - l / n wV. 

All fixed end moments are negative and are recorded in 1000 ft. lb. in the 
computation form below. The unbalanced moment, U, defined in Section 2 
as the M F to the left minus the M F to the right, equals 



(-33) - ( 
(-73) - ( 



73) = +40 at A, and 
33) = -40 at B. 



The corrections to the fixed end moment due to the unbalanced moment 
of +40 at A equals 

+2 X QU = +2 X % (+40) = +20 at A, and 
-1 X QU = -1 X M (+40) = -10 at B. 

The unbalanced moment of —40 at B gives the following corrections: 

+1 X QU = +1 X M (-40) = -10 at A, and 
-2 X QU = -2 X M (-40) = +20 at B. 

The final moments, M AB and M BA , equal the algebraic sum of M F and 
the two corrections. 



n = H n = H 
w=\000, 1=20 1 A 10=2200, f=20 1 B w='\0Q0,l=20 

\S 1/ 


-VaX1 000X20-' 


-33 


-73 


-Vi2X2200X20 2 


■ - 

-Vi2X2200X20 2 I -73 


-33 


-VijX1000X20 2 


(-33)-(-73) 


+40 


+20 


+2XKX(+40) 


-KXC+40) 


-10 










-10 


+ KXC-40) 


--2XKX(-40) 


+20 


-40 


<-73)-(-33> 






-63 


M AB 


M BA 


-63 







Maximum positive moment in span AB is 
+ H X 2200 X 20 2 - 63,000 = +110,000 - 

In this case, due to symmetry, the moment 
been simplified as follows: 



63,000 = +47,000 ft.lb. 
computations might have 



♦Maximum center moment requires (1) full live load on AB and (2) minimum restraint at joints A and 
B; minimum restraint from beams adjacent to A and B is obtained when they carry no live load; and 
minimum restraint from columns adjacent to joints A and B requires load on alternate spans as shown on 
floor above and below floor AB in Fig. 4(a), 



60,000 



Q0,000 



40. 00 




20.000- 



(b) 






us 



20 

4, 



Inferior span : A 8 



8 



Q 



i 



4n + 2 



4 



1 1 *- n 



60,000 



60. 000 



40.00 




20.000 



OS 



u^ 



20 



4i 



f 



Exterior spBn ; A 5 



i 



Q* 47T7i 



li - 



n 



n) 



Fig. 4 



rT.L=2200 ,D.l.*IOOQ 



20-0 



-i 



' A/' 



20-0 




20-0 



it 




• ^ " 



20-0 



* A U 



20-0 



20-0 




Fig. 5 



M AB = M BA = -73,000 + M X 40,000 = -63,000 ft. lb., and 
Max. pos. mom. = +110,000 - 63,000 = +47,000 ft.lb. 
To appraise the effect of varying Q-values, the end and center moments 
given below are determined by both approximate and exact analysis* for 
Q equal to V20, Y% an d /4> using the load arrangement in Fig. 4(a). 



Q 


H 


^ 


l /« 


n 


H 


m 


4M 


End Moment . ... (Exact) 


-63,000 
-60,000 


-68,000 
-65,300 


-71,000 
-69,700 


Center Moment . . . (Approx,) 


+47,000 
+50,000 


+42,000 
+44,700 


•■ 


h 39,000 
h40,300 





The moments are plotted in Fig. 4(b). It is seen that increasing Q from 
V20 to Y /± (equivalent to decreasing n from 4 J/2 to J^), decreases the end 
moments and increases the center moment by about 20 per cent. The 
greatest discrepancy between any corresponding exact and approximate 
valuo is about 6 per cent or only one-third of the disparity in moment 
when Q varies from V^ to J^. 

*"Exact Analysis" in thie text indicates that the frame in its entirety was analysed bv the moment 
distribution procedure proposed by Professor Hardy Cross (sec Reference 10). The intermediate calculations 
were recorded to the nearest 100 ft.ll>.. and slide rule was used throughout. Due to cumulate rrors, ti 
discrepancies in the "exact" values may possibly be as large as *500 ft.lb. 



10 



The center moment computed by using the conventional coefficient 
(Hi^P) * s a ^ so plotted and is about 17 per cent greater than the approximate 
moment determined for Q = M ana< 41 per cent greater for Q = V20. 

In Figs. 4(a), 5(a) and 6(a), note that two joints and loads on spans 
adjacent to these two joints only are considered in the determination of 
beam moments by the proposed procedure. It is merely for comparison of 
these moments with the exact maximum moment values that the entire 
frames and all the loads shown in Figs. 4(a), 5(a) and 6(a) are included. 

■ 

Problem 2. The frame in Fig. 5(a) has equal spans, 20 ft. long, and equal 
story heights; dead load equals 1000 lb. and live load 1200 lb. per lin. ft. 
Determine the maximum center moment in the exterior span AB and the 
corresponding end moment M A b. 

To produce the maximum value of the center moment, the loading is 
to be arranged as in Fig. 5(a).* Determine the fixed end moments accord- 
ingly. The beam and column dimensions are not known. Suppose the 
exterior columns are slender (Q A = }4) and the interior columns of average 
proportions (Q B = Y%), see Section 3. The calculations are as follows, with 
moments given in 1000 ft.lb. 



v>=0 


1 A 


w = 2200, J =20 


1 B tt> = 1000, I =20 





-73 


- l /iiX2200X20 J 


-VaX2200x20 2 


-73 


-33 


-V12XI 000X20* 


1 O-C-73) 


+73 


+37 


+2X^X<+73) 


-KX(+73) 


-18 










- 5 


+ ^X(-40) 


-2XHXC-40) 


+10 


-40 


(-73)-(-33) 






-41 


M AB 


Mba 


-81 







The moment calculations in the above table may be written in the follow- 
ing simplified manner, the computation form in Fig, 2 being omitted: 

At A: -73,000+2X^X73,000+1 X^X (-40,000) = -41,000 
Atfl: -73,000-1X^X73,000-2X^X(-40,000)« -81,000 

Average: —61,000 

The maximum center moment in span AB equals • 

+y 8 X 2200 X 20 2 - 61,000 = +49,000 ft.lb. 

To appraise the effect of varying Q A} values of end and center moments 
given below are determined by both approximate and exact analysis for 
Q A equal to M, Vs and V20; Qb remaining equal to J^. 





Qa 




H 


H 


Vto 




n A 




H 


i« 


4}^ 






. . . (Approx.) 
. . . (Exact) 


-41,000 
-41,800 


-60,000 
-60,500 


-71,000 
-71,800 




. . . (Exact) 




-49,000 
h51,000 


+44,000 
+46,000 


+41,000 
+42,500 

















♦See footnote on page 9. Note that the maximum value of Mab is not produced by the loading in Fig 
5(a), but by a loading as shown in Fig. 8(c). 



11 



The moments are plotted in Fig. 5(b). It is seen by comparing Fig. 5(b) 
and Fig. 4(b) that the center moments are nearly identical in interior and 
in exterior spans; their variation when Q varies is relatively small. Com- 
pared with the conventional moment values of 1 /iqw1 2 for interior and l /\%wl 2 
for exterior spans, the moments determined by the analyses will give a 
more economic design, especially in exterior spans. 

The end moment at the exterior column, plotted in Fig. 5(b), is more 
sensitive to variations in the value of Q than any of the other moments 
plotted in Figs. 4 and 5. Values of Q should therefore be selected with greater 
care for exterior than for interior joints. The end moments at the exterior 
column determined by analysis are smaller than those obtained by the 
use of the conventional moment coefficient of V^. 

Problem 3. Frame dimensions and loading are the same as in Problems 
1 and 2. Determine maximum end moments, M B a, in span AB at first 
interior and at interior columns. 

The loading arrangements required to produce the maximum moments 
are shown in Fig. 6(a).* The analyses that follow are carried out according 
to the procedure in Problems 1 and 2, assuming for all columns Q = J4, Y% 
and y 2 o. The results are tabulated below and plotted in Fig. 6(b). The 
end moment is slightly greater at the first interior column than at the 
other interior columns; and both moments are greater than the moments 
computed by using the conventional coefficient of Vi2. 



100,000 



60,000 



T 



lit Interior column. 



60,000 
(b) 




Interior column 



40,000- 



20,000 







5». 



^ 



Approximdte 
Exact 



Q- 



1 



4nfZ 




In ter'tor column 
I 1 



fa) 



A 



3 




1 
10 

4i 



3 

li 



— Q —+ 



' 

I if Interior column 



n 



z 



Fifj. 6 



♦Maximum end moment at B requires (1) full live load on AB, (2) maximum restraint at B. and (3) mini- 
mum restraint at A; requirement (2) calls for full live load on span adja'i nt to B and (3) calls for no live 
load on span adjacent to A. Further studies indicate the load arrangement shown above and below floor AB 
is required to give maximum moment at B, 



12 



Q 


M 


H 


VfD 


n 


H 


Di 


4H 


First Interior Column, End Moment. . . (Approx.) 

(Exact) 


-91,000 

-90,100 


-82,000 
-83,300 


-77.000 
—78,100 


(Exact) 


-83,000 
-86,500 


-78,000 

-81,700 


-75.000 
-77,600 



Problem 4- The frame and loading will be similar to those in Problems 
1, 2 and 3. The value of Q = Y%{n = V/i) will be used at all joints. The 
span in which moments are to be determined and also alternate spans 
will be 20 ft. long; for the other (intermediate) spans, lengths of 16 ft. 
and also 12 ft. will be chosen. 

In the 20-ft. spans, maximum moments will be determined at the follow- 
ing points of the beams: Center Interior, Center Exterior, End Interior, 
End First Interior and End Exterior. 

The loading arrangements required to produce the maximum moments 
are similar to those shown in Figs. 4, 5 and 6, except that the approximate 
end moment in the beam at exterior columns is produced by a loading as 
in Fig. 8(c) and determined by equation (6). All other approximate 
moments are determined according to the procedure in Problems 1 and 2, 
and the results* are summarized in Table 1. 



Main Span: 20ft 


Adjacent Span Length 


Maximum Moments 


20 ft. 


16 ft 


12 ft 


(Exact) 


+42,000 
+44.700 


+43,000 
+45,000 


+44,000 
+45.200 


(Exact) 




H44.000 
h46 t 000 


+45,000 
+46,100 


+45.000 
+46,200 


(Exact) 


-78,000 
-81.700 


-73,000 
-77,000 


-69,000 
-72,900 


(Exact) 


-82.000 
-83,300 


-76,000 
-77,900 


-70,000 
-73,800 


(Exact) 


-63,000 
-65.000 


-63,000 
-65.400 


-63,000 
-65,800 



Tabic 1 



The agreement is good between results obtained by the approximate 
and the exact analyses. The change in moments in the 20-ft. spans is small 
when the length of the adjacent spans decreases. The maximum change 
in the center moments and in the end moment at the exterior column is 
insignificant. The end moments at the interior columns are somewhat more 
sensitive to change in length of adjacent spans. 

Moments in the shorter spans will be discussed in Sections 8 and 9. 

♦The calculations may readily be duplicated by the reader by using Fig. 2. 



n 



Conclusions: For the cases investigated in Pi <lems 1, 2, 3 and 1. the 
summari. in Flj 4(b), 5(b), 6(b) and in 1 1 show that : 

(i) The agreement between momenta obtained by the approximat* ml 
by th« • ; ■ - dun is fiatifif actor 

(2) Tl ■ moment • pt the end moments at ext< rior lumna, < nj: 
comparatively little when the relative stiffness, ?.. varies;* 

(3) .Moments obt ted by using conventional ooeflictanta me % ater (up 

to about 75 per cent) than moment^ obtained I analvsifl, exoc i 

that end momenta at interior and fii>t interior columns are small* i 

(op bout 20 per cent) than th< obtained by analysis. 

lor neentr 1 Loads winch an symmetrical with reap to tin mid- 

point of th< joii, the agreement bet 1 en momenta <>bi 1 by the pio- 

p 1 ami by the i lei procedure WtJ found ^ factory. It is po.vnbl* 

however, i irranfj. load unaymmetrically with reaped to the midpoints 
pans that the p ro po s ed procedure max \i ■ unsatisfactory result- but 

such l<i:i(jinga ao rela ly infi tit. 

It was r nmended in ■Vet ion 3 that tin minimum valu< lx- el < n 



one-half. If n illy smaller than this value, using n — J*.- i •■■■• 

< r ii\ suits for thi that an especially senait <• to \ v 

narnelv. the moments in columi and in beams at ext< •■>! <<»!- 
urnm — • -• - !-'■§£ b d 1 ig 9 Other momenta are anVn far !• — I- 

isll ciinii). and, in general, pi ed no1 l»« takei imalh 

than on( -half. 

By the simplified | dun sent I, ma: tnum momenta in continuous 

buildii fran»es ma> !• letcrmined by w mg a U -u figure.' ud a m< 

al distribul m oi n tl<nt.- n\uy o|>! tied than 

b ie convi ill j.iit.' nl e< h 11 lit,*- 

M«.i at critical po dol ed b) tin ap] imat* method aj 

rail) f( f <» be than lb « a* f momi n1 ( onaidei ■ 

t hi i tluixm m I ip < On i he frame i 

s an i>t« in for < imuni momt nl hut on 

o ppn \ ,?it< moi I l.i i1 i h\ 2 apaii 

loading i i 1 |m am all* than !h< fid i monf It hh<»ul< 

t ! • • rn)>l< tr loading i r i I . I vs ill • st r 

§trt] c i nt will n 






e^Hy imium «i. and '.'< I 

t««d ) (Mttt* b| u*« 

•aw F»r* « I »vid lviu»' 
tt> aJJ tipam' A 




Em m^mtamt at e^rmiBav — - v& 



24* • 

B»d MMBte ia ***** t*mmm li- 

ft* 



*tafab work *< «panM ol arrurary b> t^iif 



'a 6*f» 



14 




reached, it is rational to use a smaller moment for design. The approximate 
moments, therefore, are good design values. 



5. Moments in Beams in Roofs and Bottom Floors 

One column connects with each joint in the roof, while there are two 
columns for each joint in floors. This difference will be reflected in the 
expression for Q in Fig. 2, for which the following equation will be 
substituted : 

which applies to beams in roofs. Otherwise, the analytical procedure illus- 
trated for floors and arranged as in Fig. 2 will apply to roofs also. 

Beams in bottom floors may be analyzed as described and illustrated 
for regular floors using the value of 

Q = — (2) 

* 4n + 2 

A part of the regular analytical work may be eliminated when the col- 
umns are relatively stubby. The calculations involving Q and U (see Fig. 2) 
may usually be omitted in the following cases: (1) at interior columns 
whose stiffness ratio n exceeds about 5, and (2) at exterior columns whose 
stiffness ratio n exceeds about 7J^. The fixed end moments may be used 
in these cases as final moments, since correction for joint rotation is 
unjustifiable. 



6. Shear in Beams 

Shear in a beam which is part of a frame is composed of the shear in the 
beam considered simply supported and a correction due to the end moments 
produced by the frame action. 

Shear in a simply supported beam may be determined by statics. For 
concentrated loads, it is usually satisfactory to use the center-to-center 
distance between supports as the span in the shear calculations. When the 
load is uniformly distributed, the shear at the face of the support is one-half 
of the load on the clear span. 

An end moment Mab, creating tension in the top of the beam AB in Fig. 
7(a) with span length I, produces reactions, R, the direction of which is 



M A A She3rCur¥e M:0 M JR [ She8rCUrve M 5A »>- dhear curve 



k 1 



M 






££/ R= —^ ZL R-- 



\L M Am *M M / 



A3 "'BA 



(a) (b) Vertical load (c) Wind load 

Fig. 7 

15 



upward at A and downward at B, the numerical value of the reaction- 
being M AB /l. The shear is constant and the shear curve is as shown in 
Fig. 7(a). The shear is also constant when moments are applied at both 
ends of the beam; but the numerator in the shear formula is the difference 
between the end moments when their direction is as shown in Fig. 7(b) 
and the sum of the end moments when their direction is as in Fig. 7(c). 
Allowance for sign of the shear in Fig. 7(b) may conveniently be made 
by use of the rule that the reaction on the beam is directed upward at the 
joint with the larger moment; the direction of the reactions in Fig. 7(a) 
and (c) may readily be determined. 

To produce maximum end shear at joint B in span AB, see Fig. 6(a), 
it is required that (1) span AB is fully loaded, (2) M BA is as large and 
(3) M AB as small as possible. This indicates the loading shown in Fig. 
6(a). For maximum end shear at exterior columns, loads may be arranged 
as in Fig. 5(a). 

For illustration, the maximum shear in the end span at the centerline 
of the first interior column in Problem 3 for n = Y2 (Q = 14) may be com- 
puted by the approximate procedure which follows, the loading being 
arranged as on the middle floor in Fig. 6(a), first interior column: 



from which 



M B a = -73,300 -lXJiX 73,300 
Mas = -73,300 + 2 X l A X 73,300, 

M AB ~ Mba = % X 73,300 = 
I 20 



Shear at B equals Y 2 X 2,200 X 20 + 2,800 - 22,000 + 2,800 = 24,800 
lb. In this case the simple beam shear, 22,000 lb., ha* been increased by 12} 
per cent at the first interior column and decreased by the same amount at 
the exterior column. One of the effects of continuity is to transfer shear; 
and it is often advisable to make allowance for shear transfer in the deter- 
mination of column loads, particularly when the exterior columns are rela- 
tively slender. 



7. Column Moments and Beam Moments at 
Wall Columns 

Maximum moments in columns are produced by placing loads on several 
beams in at least two bays. The following convenient procedure, which is 
based on placing loads on few beams only, gives results that are satisfactory. 

Fig. 8(a) shows a loading arrangement that produces, approximately, 
maximum moments in interior columns.* Live load is placed on all beams 
in one baj r and there is no live load on beams in adjacent bays. The unbal- 
anced moment, U, tends to rotate each joint through the angl< O, and 



♦Maximum moment at the cade of a column requires lhat (1) th<- unbalanced moment at t li • ends is 
maximum, (2) the column deflection curve has reverse curvature, ar '>c rotation of tin id j"inta is 

maximum. This indicates approximately the loading shown in Fi«. 8 A somewhat target tnomenl >uld 
l>f produced if live load were placed as shown on two floors on I the loading reversed on the next floor 

a! rod below. 

16 





the columns will deflect as indicated 
in Fig. 8(a). The moments, M, 
induced in the columns by U are 
shown in Fig. 8(b). 

For the conditions in Fig. 8, it 
can be shown, see Section 28, that 
the end moments in the interior 
columns equal 



M = 



3n 



Fig. 8 



Qn + 2 



X U, 



(3) 



in which 

n = ratio of K for columns to K for beams (see expression for n in Fig. 9) , 
U = unbalanced moment due to the vertical loads indicated in Fig. 8(a) . 

In exterior columns, see Fig. 8(c), end moments may be computed from 
the following equation: 



M = 



2n 



X U, . 



m * * 



in + 1 

in which U equals the fixed end moment at the exterior joint. 

The procedure is to 

(1) determine U from the loading given, 

(2) multiply U by the proper coefficient selected from Fig. 9. 



• . . (4) 



The moments thus determined 
are approximate. To be conserva- 
tive, it is advisable to choose large 
values of n and to adopt, say, n = 
l /2 as a minimum value. 

The shear, V, in columns with 
reverse curvature as in Fig. 8 is the 
sum of the end moments divided by 
the story height, or in this case 



V = 



2M_ 



(5) 



50 



^30 

^> 

45/0 

I 







Ext 


II \ 2) 

er/or columns- dn 


















v &n 


'nteriorcolumns 






















I 




















It 








































i 

















































































12 3 4 

Relative column stiffness 

Fig. 9 



8. 



Since the end shears are opposite 
and nearly equal in the column 
above and the column below a joint, 
the beam between the columns gets 
little or no axial compression from the loading in Fi 

Maximum end moment in beams at exterior columns equals the sum of 
maximum moments in the column above and the column below; it is 
therefore produced by a loading similar to that which gives maximum col- 
umn moment and may be taken as 

M AB = —^- X U, (6) 

4n + 1 

in which U is the fixed end moment, M^b- 



17 



Problem 6. A frame has spans and loading as in Problems 1 to 3; values 
of l A, l l A, tyb are assumed for n. Determine maximum moments in the 
columns and in the beam at the exterior columns. For the loading arrange- 
ment in Figs. 8(a) and 8(c), the unbalanced moment, U } is 

at interior cols. : V12 X (2200 - 1000) X 20 2 = 40,000 ft.lb. 
at exterior cols.: l / n X 2200 X 20 2 = 73,300 ft.lb. 

Using equations (3) and (4), the following values are computed: 



n 



Interior Columns 



(Coefficient) 
(Moment) 



Exterior Columns 



(Coefficient) 
(Moment) 



\4 



1H 



0.300 
12,000 



0.333 
24.400 



0.410 
1 6,400 



4H 



0.465 
18,600 



0.429 
31 ,500 



0.474 

34,700 



Stresses due to these column moments are illustrated in the following 
examples: 

(a) Interior column, n = 4J^; gross section = d 2 = 24 X 24 in.; axial 
load, P = 500,000 lb. The eccentricity is 



M 12 X 18,600 n ._ . 
e = — = ■ = 0.45 m 

P 500,000 



(b) Exterior column, n = 1}4', gross section = d 2 = 16 X 16 in.; axial 
load, P = 200,000 lb. The eccentricity is 



e 



M 12X31,500 tn . 
= — = - = 1.9 m 

P 200,000 



In a rectangular, homogeneous section with depth d, the extreme fiber 
stress for load P with eccentricity e is 1 1 + — J times the stress produced 
by load P applied concentrically. The increase in stress is 

for (a): '— = 0.11, or 11 per cent, 

w 24 

f or (b); 6 X L9 = 0.71, or 71 per cent. 

16 

In this example, the effect of moment is negligible in the interior column ; 
but the moment in the exterior column should not be ignored. 

The maximum moments in the beam at the exterior column are, using 
equation (6) for the approximate values: 



n 




Moment 



(Approx.) 
(Exact) 



M 



1H 



0.857 



4H 



0.948 



-48,900 
-49,900 



62,800 
64,700 



69,500 
70,500 



There is good agreement between the approximate and the exact moments 



18 



Special Procedure of Analysis 

8. Assumptions Regarding Restraint at Far Ends 
of Columns 

To obtain satisfactory results by the procedure in Fig. 2, the rotation 
must be relatively small at joints in the floor above and below the floor 
considered. If the rotations are large, a considerable error may result from 
assuming that the far ends of the columns are fixed or hinged. The errors 
that may be large are in the moments that depend upon live load being 
placed in short spans which are flanked by long spans without live load, this 
loading distribution being reversed in the floor above and below. In general, 
such moments are relatively small and not important; but a discussion of 
the problems involved is desirable and will be given in the numerical 
problems that follow. 

Problem 6. The symmetrical frame in Fig. 10 has three bays, the span 
lengths being 20 ft. for the outer spans and 10 ft. for the center span. The 
dead load is 1200 and the live load 2400 lb. per lin. ft. All Jv-values will 
be assumed equal, and the value of n is therefore equal to unity. In order 
to produce (1) minimum center moment in AB, and (2) maximum center 
moment in BC, live load is placed on the short beam BC, the general loading 
being arranged as in Fig. 10(a). 

The actual condition of continuity is shown in Fig. 10(a), and an assumed 
condition with the far ends of the columns fixed is illustrated in Fig. 10(b). 




fa) Actual condition of continuity 

A 3 C 



3 



6 



-dO.O +20.0 -40.0 



-28.2 + 24.8 -42.2 
-25.5+25.9 -42.6 



-59.9*16,5 -47.2 



-28.4+24.1 -43.4 



-30.0 +150 



-32.2 + 12.8 

-33.1 + 113 



-25.1 +19.9 



-31.6 +13.4 



-35.6 +/ 7.7 -48.9-25.5+19.5 



A 



(c) 



3 



nillllllllllllTTT 




^ » 5 



n 



IHIHMMI 




(b)Far ends of columns assumed fixed 



"*"» ** 



»4 



//% ■ Actual as In (a ) 



■V •* 



T&6 



Proposed procedure 
fixed enos as in fb) 



C 

Fig. 10 



w 



19 



The deflection of the centerlines is indicated by the dotted lines. It is seen 
that the columns deflect in opposite directions and that the rotations of 
joints B and C are opposite in the two figures; this indicates that the 
assumptions made in Fig. 10(b) are unsatisfactory. 

Moments in the beams AB and BC calculated on basis of various assump- 
tions are tabulated in Fig. 10(c). Line 1 gives the moments obtained when 
all joints A, B, C and D are fixed. When these joints are released and the 
far ends of the columns remain fixed, as indicated in Fig. 10(b), the 
moments are as in Line 2; Line 3 gives the moments obtained when the 
columns are assumed hinged at joints above and below ABCD. Compare 
the moments in 2 and 3 with the exact moments, which are given in 4, 
and with the moments in 5, which are obtained by the procedure in Fig. 2. 
The results recorded in lines 2, 3 and 5 are similar; and they are in error, 
obviously because the joint conditions assumed at the far ends of the col- 
umns are not correct. 

For loading and frame conditions similar to those in Fig. 10(a), too 
great an error may result from assuming the far ends of the columns fixed, 
hinged or partly restrained. In such cases, proper allowance should be made 
for the effect of loading on the floor above and below the floor considered. 
The allowance, which is made in the moments recorded in Line 6, will 
be discussed in Section 9. 



9. Correction for Rotation at Far Ends of Columns 

In making allowance for rotation of joints above and below floor ABCD 
in Fig. 10(a), the analytical procedure in Fig. 2 is useful but may be modi- 
fied as illustrated in Problem 7. 

Problem 7. The frame, loading and quantities sought are the same as in 
Problem 6. The calculations, given first without correction for joint rotation 
above and below ABCD, are recorded in Table 2, which is arranged as 
Fig. 2. The end moments determined in Table 2 — Mab, Mba and Mbc 
— are those 1 given in Line 5 in Fig. 10(c). 



B 

I 



y 6 



40 



-Vu XI 200X20' 



I, 



-V.sX1200X20 s 



+13 3 



- 1.7 



+2X'«,X(+40) 
+1X'6X(-10) 



-1XJcX(+40) 



-2X'fcX(-10) 



-28 4 



M 



m 



Mh 



-40 



-67 



+ 3 3 



-43.4 



-30 



-33 



+ 1.7 



/nX3600X10' 



+2 X Hi - 1 0) 



+1XH(+10) 



-31.6 



.1/ 



BC 



Table 2 

The values of 40 and 10, underscored in Table 2, are the unbalanced 
moments at 4, B and C. They are the only values which will be corrected; 
the procedure of correction is illustrated in the paragraphs which follow. 

The numerical value of a moment at one end, A, of column A A' induced 
by a rotation due to an unbalanced moment, V a', at the other end, A', 
will be taken as n X l\\> X Qa' (see Fig. 11). The sum of moments origi- 



20 




U.-=(-l?QO) 
nU A .Q A .'l(tl?0.0)*i" 10. n U.. Q..'lflJ0.0)'i'-l6.3 








W--I700 



nU.. Q.il(*1200)*i' 1 70. 




-J-l?O0h' 120.0 




nU 6 ,Q r -l(-IIO.O)4-.-m 

(/*.*(■ naoH-il 



-no. o 



»,3bOO 



70-0 




w-noo 



Q'( a+ ail joints 

Fig. 11 



nating at the joint above and the 
joint below A, equalling nU A 'Qa' + 
nU A >>Q A " f tends to rotate joint A in 
a direction opposite to the direction 
of Ua' and U A ». This sum will be 
included in the determination of the 
unbalanced moment at A, thereby 
introducing an allowance for rota- 
tion at A' and A" . 

The procedure is illustrated in 
Fig. 11. The fixed end moment at 
A' and A" is -120.0, and U r = 
U A » = - (-120.0) = +120.0. 
Choosing the value of Q A > = Q A „ = 
% and inserting n — 1 gives the 
sum of the moments induced in the 
column at A : 

1( + 120.0)^ + 1 ( + 120.0) Jf = 2(+20.0), 

the direction of which is opposite to the direction of M AB . 
The moment tending to rotate A then equals 

U A = - (-40.0) - 2(+20.0) = . 

Similarly at column B'BB'\ taking Q B . = Q B „ = % gives 
U B > = V B » = -120.0 - (-10.0) = -110.0, and 
nU B 'Q B > + nU B »Q B » = l(-HO.O)}^ + 1(-110)^ = 2(-18.3). 

The direction of the moments induced in the columns is opposite to that 

of M BA , which gives 

U B = -40.0 - (-30.0) - 2(-18.3) = +26.6. 
In summary, to obtain a corrected value of U at any joint, take the 
M F to the left minus the M F to the right of the joint minus the moments 
induced in the columns above and below the joint. The moments induced at 
the near end of the columns may be taken as nQU written for the far end 
of the columns, U in this case also being equal to M F to the left minus 
M F to the right. 

In the moment calculations in Table 2, replace +40.0 by 0.0, — 10.0 

by + 26.6 , and +10.0 by —26.6. The calculations will then be as given 

in the following table: 



k 



-40.0 







_i 



/iiX1200X20= 



B 

I 



_i 



/12XI 200X20! -40 



+2XMXC0) 



4- 4.4 



-35.6 



+1XMX(+26.6) 



1/ 



AB 



-IXHX(O) 



-2XJ^X(+26.6) 



M 



/*. 1 



0.0 



- 8.9 



-48.9 



-30 



_i 



, uX3600X10 2 



+ 89 



- 4.4 



+2XK(+26.6) 



-HX!r,X(-26.6) 



-25.5 



.1/ 






The center moments equal : 
in AB: y s X 1200 X 20 2 
in BC: H X 3600 X 10 2 



M(35,600 + 48,900) = 17,700 ft.lb 
25,500 = 19,500 ft.lb. 



21 



These moments are recorded in Line 6 of Fig. 10(c) and plotted in curve 
marked "6" in Fig. 10(d), in which they are compared with the exact 
moments (marked "4") and with the moments marked "2" which are 
obtained by assuming that the far ends of columns are fixed. It is seen by 
comparison of the moment curves in Fig. 10(d) that the assumption of 
fixed far ends is unsatisfactory in this case, and that the proposed pro- 
cedure gives moments that are close to the exact moments. 

It should be noted that for a frame as in Fig. 10(a), the general procedure 
in Fig. 2 — without correction — gives good results for the moments in the 
floor above and below ABCD. For further illustration, the following cal- 
culations and comparisons of moments applying to the frame and loading 
as above are presented for two conditions: (1) all i£-values equal to unity, 
Q = }4 (moments underscored); and (2) all i£- values equal to unit, Q = %f 
except for the intermediate beams, for which K = 1/10 and Q — 1/42 
(moments given in parentheses). All moments are in ft. lb. 

Maximum center moment in exterior span. 

At exterior column: 

-120,000 + 2 X H X 120,000 + 1 X K(-H0,000) = -98,300 
At interior column: 

-120,000 - 1 XK X 120,000 - 2 X V&{- 110,000) = -103,300 

Average: = -100,800 

At midpoint of span: 

Case (1) Case (2) 

Approximate analysis: +180,000 - 100,800 = + 79,200 (+79,200) 
Exact analysis : + 82,000 (+84,700) 

Minimum center moment in interior span. 

Case (1) Case (2) 
Approximate analysis: 

+ 15,000 - 10,000 - % X 110,000 = - 13,300 (+2,400) 

Exact analysis: - 15,600 (+2,500) 

Maximum end moment in beam at exterior column. 

Case (1) Case (2) 

Equation (6): ^^ — - X 120,000 = -96,000 (-96,000) 

H w 4X1+1 — 

Exact analysis : - 102,700 ( - 103,800) 

The center moment of —13,300 ft. lb. under the heading "Minimum 
center moment in interior span" is produced by the loading shown on the 
top or the bottom floor in Fig. 10(a). For the same loading, the end moment 
in the interior span is 

-10,000 - Ve X 110,000 = -28,300 ft.lb. 
Since moments created by this loading are negative throughout the entire 
length of the center span, it will be necessary to detail top bars extending 
from span AB through BC into CD. 

The practice of stopping top bars at or near the quarter-points of spans 
is safe where adjacent spans are nearly equal and live load is relatively 
mall. When the difference between the lengths of adjacent spans inrreasi 

22 



and/or the ratio of live to dead load increases, it becomes increasingly 
important to determine the points to which top bars should be extended, 
points which may be quite different from the quarter-points. It is recom- 
mended to arrange the load as for example on the top floor in Fig. 10(a), 
to compute the corresponding moments ( — 13,300 and —28,300 in the 
example), and to sketch the moment curve connecting these points. Such 
curves indicating minimum center moments are of importance in design 
and detailing of bar reinforcement. 

In the cases illustrated, it is seen that (1) the accuracy obtained by 
use of Fig. 2 and equation (6) is satisfactory, and that (2) the change of K 
for the interior beams from unity to one-tenth has comparatively little 
effect upon the moments beyond the joints between which the change 
takes place. 

10. Summary 

The following procedure is recommended for determination of moments 
produced by live load placed in short spans flanked by one or two long 
spans without live load. Arrange and carry out the calculations as in Fig. 2 
with the exception that the unbalanced moments are modified in accordance 
with the following steps, (1) to (5), which are illustrated in Fig. 11. 

(1) Determine Fixed End Moments (recorded to right and left of all 
joints). 

(2) Compute Unbalanced Moments, U, at joints A f , A", B', B" (on 
larger circles) . 

(3) Determine Q at joints A\ A", B', B" . 

(4) Compute nQU at A', A", B', B" (recorded above and below A 
and B). 

(5) Compute Modified Unbalanced Moment at A and B, to be used in 
Fig. 2, as M F to the left minus M F to the right, minus the moments 
induced in the column above and the column below each joint. 

The calculations as arranged in Fig. 11 are simple and direct; they may 
usually be recorded without the schematic arrangement. 

An outline of the analysis for vertical loading in accordance with the 
general and the special procedures is given below together with references 
to sections of the text where derivation and description are presented. 



Vertical Load 

i 



,1 



General Procedure 
I 



Maximum Beam Moments 

I 



End Span at 
Ext. Cols, 



M = 



4n 



4n+1 
(Sec. 7) 



X U 



All Others 



1 



Q 4n+2 

and Fig, 2 

(Sec. 2) 



I 

Special Procedure 

I 



Maximum Column Moments 

I 



I 

Exterior 



M = 



2n 



4n+1 
(Sec. 7) 



X U 



Interior 



M = 



3n 



6n+2 
(Sec. 7) 



X U 



Beam Moments 

i 



i 



Same as general pro- 
cedure but with allow- 
ance for rotation at far 
ends of cols. (Sec. 9) 



23 



Dim f/vwon <>/ I arums / //# < [$ 



\ fr:t! • I m -i die! «i n» t* 

1 1 all 1 li< [•ml : « w idl l»> i luti nft 

m:|UU ill 1 Ur tlih 

|>l ! \S i Loll £1\ V \ Mil | |i>< 

MUli, I .' U lid t mtf In 

lid oth - I In fen | »)t I It 

.1 1 li «jl I : 1 1 1 < i I |f 111 



1.. 









* «»f lit 



f fmi 



in 



j 



< ai 



illH 






I 






• ■ I * 



I ll< ll|l lilt I J 

I ip! ii I ip 

1 1 } I ( i Mil ' i 

i • Mi> I !i(KMk lit* I 






•* f 1 I I 

1 
I 



n a1 •« pinn :\ - w I 

it i ix unknown 

liii< i. 

I ; i . t' a* 

i|» « lit fMiii v\ li * dc rd i 



i 



p 









u 



\ 



I m I 






hi 



' 












'1 M 



li imp! 

ii y I ) 1< .ii at 

.id tin nl hi 

ii u\> < < ! i d 1« r i !1i « i «ll I. nl 






i 



,i 












I s 









. 



f ro) mi 



liili 












H 



r< m , // I n it, u if |( I 1 liftflf 



. i 



f < 






\ ll 



i i 



J l 






*+ 














■ 









j 




i* i» 



comparison with moments obtained under certain typical conditions (a), 
(b) and (c), each case illustrating the corresponding point given above. 

(a) Assume that the column reaction is triangularly distributed as shown 
in Fig. 12. The area of the triangle, or the reaction component, is 

VA 2— J- = %V ' y and the distance from the center of gravity of the triangle 
to the column centerline is %( - ) = -, The moment of the component with 

respect to the column centerline is J^Vf r J = x /§Va. This moment, which 

equals the reduction proposed in point (a), is approximately the amount 
to be deducted from the centerline moment, M c , when a reaction concen- 
trated in the column centerline is changed to the triangular reaction shown 
in Fig. 12. 

(b) Consider two fixed end beams with the same load, w lb. per lin. ft., 
the span being I in one, but (I — a) in the other beam. The difference between 
the end moments in these beams is 

Vi2^ 2 — \ nw(l — a) 2 = y$wla — l /nwa^. 
The term ^Inwa 1 is small compared with ' () u-la; by disregarding it and 
substituting V for l^wl, the difference between the two fixed end moments 

becomes %Va, which equals the correction recommi oded for the moment 
at face of column. 

(c) The slope of the tangent, bf, to the moment curve at b in Fig. 12 

equals V and the distance be equals Vl J. The modified momenl curve, 

which is shown dotted in Fig. 12, lies above the original moment curve 
between the column faces, the distance between the curves being somewhat 

less than MF(-) = Y§V a , sav > %( l A>^' n ) = ' aF«. This is the reduction 

of the center moments recommended due to effect of width of column. 
In Problem 1, the positive center moment, 47,000 ft. lb. was determined 
for w = 2200 lb. per lin. ft., and / = 20 ft. assumed to be the distance 
between the centerlines of columns. Making allowance for a column width 
of, say, 1 ft. 6 in., the corrected center moment is reduced to 

47,000 - r g(2200 X ^°)l-5 = 47,000 - 3,700 = 4: J, 300 ft. lb. 

The loading in Fig. 6(a) for maximum end moment, 91,000 ft. lb., at 
the first interior column, B, produces at B an end shear of V = 24.800 lb., 
computed in Section 6. With a column width of 1 ft. 6 in., the moment 
at the face of the column at B is reduced to 

91,000 - H X 24,800 X 1.5 = 78,600 ft. lb. 

The question arises whether the beam to be designed is governed by 
the moment at the face or by the moment at the centerline of the column 
For illustration, refer to Fig. 13 in which a stress distribution is indicated 
at the faces of the column. The stress T and a considerable part of the 
stress C will be transmitted across the column by the reinforcement, but 

25 



the remaining stresses are distributed over an effective depth which is 
greater at the centerline than at the face. In European practice, the center- 
line depth is commonly taken as d -J- -, and the stress distribution is 
assumed to be as indicated in Fig. 13. The relative increase in effective 

depth at the centerline is then — -, whereas the relative increase in moment 

6a 

according to Fig. 12 is 

H Va H X V%wla a 



Mc-VsVa l / l2 wP - l AX Hwla I - 2a 

The tensile stresses would be greater at the centerline in cases where the 
increase in moment exceeds the increase in effective depth; that is, when 

a ^ a , ^ I — 2a 

> — or a > 



I - 2a 6d 6 

In general, d is smaller than (I — 2a) /6; and it is then the moment at the 
column face that governs the design. 



12. Effect of Change in Moment of Inertia 

The research made 1 on the effect of moment of inertia, /, of members 
in continuous frames, is not sufficiently comprehensive, and some discrep- 
ancy, therefore, may exist in recommendations for J taken from various 
sources. In current practice, certain general procedures have been developed 
which will be briefly discussed. 

The ratio of 

K c h X I 



n = 



Kb lb X h r 



in which c refers 10 columns and b to beams, enters into all analyses of 
building frames. Considerable uncertaintj r exists in determining ratios of 
Ic (for columns) to lb (for beams). 

In typical building frames, the entire cross-sectional area of columns i 
usually effective in taking stresses, and values of / for columns should 
preferably be based upon the gross concrete section and computed as 6d s /12, 
in which b denotes width and d depth. It is frequently recommended to 
add hereto the value of the moment of inertia, I Bt contributed by tin 
longitudinal bar reinforcement, making the total moment of inertia 



(I - •) 



If the column is square and the bars are arranged in acircl<\ tin moment 
of inertia of the column may be written as: 

/ = % 2 X d* + %a(^ - 1V1 (7) 

in which 

d = side dimension of square column, 

26 



A = area of longitudinal bars, 



Es 

Er 



= diameter of circle through centers of bars, 

= ratio of moduli of elasticity of steel and concrete. 



The area, A, in equation (7) is assumed uniformly distributed along the 
circle with diameter c; and by writing the transformation factor for the 

bars as — 1, it has been taken into account that the area A has been 

included as concrete in the first term in equation (7). 

Cross-sections in beams have both compressive stress and tensile stress. 
The question arises whether the tensile stresses in the concrete should be 
ignored. In frame analysis, I is introduced for the purpose of determining 
slopes and deflections. It is not the particular value of / at any one cross- 
section, but the /-values over the entire length between joints which govern 
deflection. Since the uncracked condition generally prevails, the gross area 
is considered to form the better basis for calculation of I. 

fed 3 
The value of 7 of a T-section is usually taken as - — , in which d is the 

total depth of the web. The lower limit of b for T-beams is the width of 
the web and the upper limit is the average value of the panel widths adja- 
cent to the web; but it is improbable that b can be equal to either one of 
these limiting values. Professor Cross* has suggested that b be taken as 
the web width, b r , times a multiplier, C, thus allowing for the effect of 
the flange. In other words, the moment of inertia of any T-beam may be 
written as 

I = CX l-^-l • . (8) 



12 

in which the quantity in parenthesis equals the moment of inertia of the 
rectangular web section. For rectangular beams, C therefore equals unity. 

Values of C for T-beams, computed on basis of the gross (uncracked) 
sections, are listed in Table 3. The values of 2, 5, 10 and 20 are ratios of 
"Flange Width" to "Web Width," and the values of 0.2, 0.3, 0.4 and 0.5 
are ratios of "Flange Thickness" to "Web Depth." 









Flange Width 




Values of C 


Web Width 


1 2 


5 


10 


20 


Flange Thickness 
Web Depth 


0.2 


1.3 


19 


2 3 


2 7 


0.3 


1.4 


1.9 


2.3 


2 7 


0.4 


1.4 


1.9 


2.4 


t 


0.5 


1.4 


2 | t 


t 


Average 


M 


1,9 


2 3 


2 7 



Table 3, Values of C 



*See Reference 6. 

tThese conditions seldom exist in typical designs. 



27 



Table 3 covers practically the entire range of flange conditions in typical 
building frames; and yet, the variation in the value of C is surprisingly 
small. When the flange width is an uncertain quantity, it is justifiable to 
consider C a constant, and the value of C = 2 is a good average. Inserting 
C = 2 in equation (8) gives the general equation for T'-beams: 

I = y 6 X 6'd», (9) 

in wmcn j _ momen t. f inertia of JT-beam 

b' = width of web 

d = total depth of web 

If the ratio of flange width, b, to web width, b f , is known, equation (8) 
is recommended for use with the average values for C given in the last 
line of Table 3. Equations (8) and (9) may be used directly to obtain values 
of /, since it is seldom considered justifiable to make any allowance for 
the reinforcement in beams. 

When there is more than one beam in a panel for each column, souk 
designers compute the moment of inertia as the sum of /-values taken 
according to equations (8) or (9), each term covering one beam web. Not 
more than one of the parallel beams frames into each column; but — it is 
contended — the other beams frame into a girder which, in turn, frame 
into (he column. In this way, all parallel beams in the panel must affect 
the relative stiffness between the column and the floor construction. It 
seems more reasonable, however, to include a fraction only of the /-values 
for beams not framing directly into the columns. 

If cross-sections are not known when the analysis begins, the designer 

has to estimate the value of n. It is then important to ascertain how much 

a discrepancy in n affects the results. 

Studies of the effect of varying the values of n are given for certain 
numerical cases in Section 1 Summaries of the results are given in Figs. 4, 
.") and 6, in which moments are plotted as ordinate*, and the a I issas 
represent values of n from Yi to V •,. a range which includes most prac- 
tical cases. The moments that change the most with varying values of n 

those at the end of the beam at the exterior column. They decreas< 

from 71,800 for // = \y 2 to 41,800 for n = Yi. Kven in this ca a changi 
in ti from 4J/<2 to Yl ( i" the ratio of 9:1 > will decn the moment in the 

rat in of about 5:3 only. 

For all other moments in Figs. 4, 5 and 6, the effect of varying the value; 

of // is surprisingly small. It is seen by inspection thai the moment! 
imputed in Sect ion 4 and given in the accompanying table an not great l.\ 

affected by relatively large variations in the ratio of f r /Ih 






■ n x >> 


*H 


M 


EnH Moment 


(Interior Column) 
(First Interior Column) 


77,600 (100) 
78,100 (100) 


86,500(111) 
90,100(115) 


Center Moment 


(Interior Span), 

(Exterior Span) 1 


40.300(100) 
42,500(100) 


60,000(124) 

MOO0 (WO) 



V •■ moments in th< table, the change with varying value of u ianc 

from 11 to 24 per nt 






13, Effect of Distant Loads 



In Fig. 2, only a small part of the entire frame is considered in the 
analysis; this makes the procedure especially convenient to apply. The fact 
that the analysis may be applied to part of a frame of such limited extent 
is in a measure due to the smallness of the effect of distant loads. 

For illustration, let the vertical 
and horizontal lines in Fig. 14 repre- 
sent a building frame in which joint 
A is rotated by an unbalanced 
exterior moment of 100. No other 
load will be considered on the frame ; 
and symmetry is assumed with 
respect to joint A. The unbalanced 
moment at .4 will induce moments 
in the frame, and the end moments 
induced at the points immediately 
to the left of each joint on the hori- 
zontal and vertical lines through A 
are plotted in Fig. 14 for values of 
n = Yi^Vi an d 43^. In the horizon- 
tal line through A, the end moments 
induced for n = 3 2 are -32.5, +5.7, —1.1 and +0.2 at distances from A 
of 0, 1, 2 and 3 times the span length. Adjacent to the column through 
A, the end moments induced for n = 3^ are —32.5, +3.1, —0.3 and 
0.0 at distances of 0, 1, 2 and 3 times the story height. The induced end 
moments become smaller when n increases; they approach zero when n 
approaches infinity (fixed end beams). 

At a certain joint in a frame as in Fig. 14, the moments induced by 
unbalanced moments two or more spans or stories from the joint are small. 
The rapid rate of decrease of induced moment with increasing distance to 
point of application of unbalanced moment is a contributing reason why 
good accuracy may be obtained with a procedure as that in Fig. 2. 




-32.S 



Fig. 14 



14, Effect of Haunching 

Haunching of beams will affect the fixed end moments produced by the 
loading and also the stiffness, which for prismatic beams is expressed as 
I/l. When analyzing frames for horizontal loads such as wind pressure, 
the stiffness, S, should be corrected for haunching by multiplying it with 
a factor which may be selected from Fig. 15 in accordance with the shape 
of the haunches. 

For example, let D/d = 2 and a = 0.3 for both ends of a beam AB 
as shown in the figure inserted in the left side of Fig. 15. The adjusted 
value of the stiffness S, taken from the left chart in Fig. 15 for D/d = 2 
and a = 0.3, equals 2.75(1 d/l), or the stiffness of this haunched beam is 
2.75 times the stiffness of a prismatic beam with moment of inertia — Id 
and span length = /. 



29 




Fig. 15 



For vertical loads, the value of I /I need seldom be corrected for haunch- 
ing in practical design problems; but the fixed end moments should be 
adjusted to allow for haunching. For the adjustment, Professor Cross has 
found it surprisingly satisfactory to use the following simple approximation. 
For a symmetrically haunched beam with dimensions as shown in 
Fig. 16(a), the fixed end moment may be taken as (1 + ab) times the fixed 
end moment in a prismatic beam with the same span and loading. When 
i he beam is haunched at one end only, the mub lier may be taken as 
1 + 2ab) at the haunched end and (1 - ab) at the prismatic end.* Tb- 
effects of haunching both ends are additive. 

Problem S. A am AB with fixed ends has unsymmetrical baun< 
as shown in Fig. 16(b) and carries a uniformly distributed load, w. Deter- 
mine the fixed end moments. 

Using the products of ab computed in Fig. 16(b) and the multiple 
f 1 + 2ab) and (1 — ab), the fixed <nd moments are adjusted for haunch- 
ing as follows: 

at A : 

( J i2<//- X (1 + 2 X %) X (1 - 1 X 3/40) = 0.116m/2 = 1.39 X x / n wP, 

at B: 

C/im'P) x ii - i x X) x ' l + 2 x m = 0.072W2 = o x VW 2 . 

• to * Pftft 07» | miifif rou* GOtnpftliMOB t 

ar-t ime&U f<yr tmth «tra L and paral I b«*. 



exart 



weao ai>i>T(jxirr-r . 



.0 



The moment is 39 per cent greater at A and 14 per cent smaller at B 
than the moment, wl 2 /12, for a prismatic beam with the same load and 
span. The effect of haunching is considerable in this example and should 
not be ignored. Moments computed by the Column Analogy Method* are 

0.115toP at A. 
O.QSOwP at B. 





I 



-3*1 




3*b=i * f=£ 



(a) 



a*b:fi*i--£ 



(b) 



Fig. 16 



WIND PRESSURE 



15. Introduction 



A single vertical frame subject to lateral load may be analyzed with a 
high degree of accuracy by procedures such as the slope-deflection method. 
In buildings, however, a number of such vertical frames — bents — act 
together, and the distribution of load to each frame must be determined 
before the analysis can proceed. This distribution is as important as the 
subsequent analysis. 

One of the current approximate methods of analysis is based upon the 
assumptions that the point of contraflexure is at midpoint of each member 
and that vertical shears in beams are equal.** These assumptions alone give 
no indication as to the amount of load carried by each bent. The assumption 
of contraflexure at midpoints becomes very useful, however, when combined 
with the assumptions presented in Appendix B. 

Bents subject to wind pressure have (1) shear deflection which is caused 
by bending in individual members, and (2) moment deflection, which is 
due to extension — shortening or lengthening — of column members. Shear 
deflection is predominant except in tall and narrow, tower-like structures. 
Moment deflection is ignored in most methods of analysis*** and is also 
disregarded in the analytical procedure in this text. It should be noted, 
however, that it is a source of error to ignore column extension, and the 
error may offset the accuracy of elaborate methods of analysis. 

♦See Reference 8. 

**First put in print in "Steel Construction" by H. J. Burt. Similar assumptions form the basis of R. Flem- 
ing's "Portal Method" applying to wind pressure analysis, see page 111, Reference 26. 

"""♦Regarding allowance for moment deflection in tall buildings and towers, see Reference 28. 



31 



A procedure is presented for determination of shears and moments in 
buildings subject to lateral loading. The derivation — given in Appendix B 
— is based upon assumptions originally suggested by Professors Wilson and 
Maney.* The method of distributing wind pressure to the individual bents 
in this text is similar to one presented by Mr. Albert Smith.** Methods 
of analyzing building frames for wind pressure will be illustrated and dis- 
cussed in the numerical examples which follow. 

16, Illustrative Problem No. 9 

Let Fig. 17 be the framing plan for a floor twenty stories below (he roof, 
the height of each story being 10 ft. Let the direction of the wind be East- 
West and its intensity 20 p.s.f. All bays are 20 ft. long. The relative values 
of /, moment of inertia, and K will be taken as follows: 

I: Relative Value K: Ratio of / 



Spandrel beams*** 
Interior beams . . 
Wall columns*** . 
Interior columns . 



of Moment of 


to Span Lengt h 


Inertia 


or Story Height 


20 


20/20 = 1 


30 


30/20 =1.5 


40 


40/10 = 4 


80 


80/10 = 8 



© f 

© 4 



® :! 



© l» 




* 



I 



,i— 





N ~* 






i ii 



=M-HI 






^ 
^ 

5 



II 
II 



II 






ii 

II 
II 

ft 



— - fr — -a^ == A.__„ 



© 



© 



© 



© 



ii 
ii 

it 

! 



I 

ii 



li 



! 



© © 



© 



© 



© 




9.5ff.y ! — J 

i W ma pressure component 

Fig. 17 



tmnine the distribution of the wind pr< e to tl columns at I he 

M<mi shon D in Fig. 17. 

-. . |,:, J] ,r B 21. 

<ff KclrrenceB >. 

*** \n B ljuBt it will !»♦• r r in tin.- viilm I / and h for bent J 






The nine frames from A to J in the East- West direction will assist in 
resisting the total wind pressure above the floor, which, with 8 bays 20 ft. 
long and 20 stories 10 ft. high equals 

W = (8 X 20) X (20 X 10) X 20 = 640,000 lb. 

Each column in Fig. 17 is part of a frame extending East- West and will 
carry wind shear. The proportion of the total wind pressure taken by each 
column is proportional to a "joint coefficient" which is a function of the 
^-values of members in the bent. The JC-values must be known or selected 
before the analysis can proceed. According to the assumption and the 
derivation in Appendix B, the coefficient at any joint may be determined 
as follows: 

Joint Coefficient = (K for col.) X ^ S " m <* *'« for adjacent beams 

Sum of K's for adjacent members 

For the floor in Fig. 17 and the iC-values given above, joint coefficients 
computed by equation (10) are recorded in the second column in Table 4. 
At joint C3, for example, the i^-value is 8 for the columns, 1.5 for the 
beam between C2 and C3, and zero between C3 and C4. The coefficient 
at C3 equals 

8 X L5 + ° = 0.69 . 

1.5+0+8+8 

The next step is to compute the sum of the coefficients for each bent 
at the floor, a sum that represents the force with which the bent resists 
a unit translation. The center of gravity of these "bent forces" is then 
computed by multiplying each force by its distance from bent J and divid- 
ing the sum of these products by the sum of the forces. The distance from 
bent J to the center of gravity equals 3458/38.61 = 89.5 ft., the values 
in parentheses being used for bent J (see Table 4). 

The distance from bent J is 89.5 ft. to the center of gravity of the bent 
forces, but 80.0 ft. to the center of gravity of the wind pressure. An eccen- 
tricity of 9.5 ft. exists wl rich tends to twist the entire building frame with 
respect to some vertical axis. The twisting or torsional moment may be 
eliminated by determining, by trial, new X-values for bent J of such mag- 
nitude that the eccentricity becomes negligible. Bent J is chosen in this 
case because it is farthest from the center of gravity, thereby giving the 
change in J great weight; and bent J may be assumed to be a wall in which 
architectural design may readily be adjusted to suit structural demands. 
In Table 4, X-values have been trebled for all members in bent J*. By 
this change, the distance to the center of gravity is reduced from 89.5 ft. 
to 3458/43.59 = 79.4 ft,, and the eccentricity reduced from 9.5 ft. to 0.6 ft. 

By adjusting the J-bent and making the eccentricity negligible, the 
advantage is gained that all joints in the floor get the same translation. 
This means that the column shears, V A , are proportional to the joint coeffi- 



> 



*By stiffening bent J, the beam shear in J (see Table 4) is made twice as large as that in the adjacent 
bents. The uplift on the windward column, Jl, is therefore twice that on the windward columns in adjacent 
bents. It is important to ascertain that the uplift in column Jl is not too large compared with the dead 
load available to counteract it. In case the uplift is too large, the stiffness of members must be adequately 
decreased in bent J and increased commensurately in the adjacent bents. 

3-3 



cients in the second column of Table 4, the proportionality factor being 

equal to 

640,000 , , ^ a 
— ! = 14,i00, 

43.59 

in which the numerator is the total wind pressure and the denominator 
the sum of all joint coefficients. 



At 



A2 



Joint 
Coefficient 



4* 



1+1 



0*1+4*4 



-44 



4' 



1*1 



1+1+4+ d 



--.BO 



A3 



A4 



A5 



AG 



42 



.80 



Columns 



Shear 
lb- 



Moment 
ft. lb. 



6,500 32, SO 



it, 600 



same 



A2 .80 



A? .60 



At 44 



4.08 



51 



32 



33 



34 



0*1.5 ri 

4* „ ,r. f,63 



8" 



0+15+4+4 
Tftt 



1 



15 +15+8-8 



ft.26 



82 1. 26 



same 



same 



Same 



59, 000 



asjoint 



asjoint 



asjoint 



asjoint 



Seams 



Moment 
ft lb. 



65, 000 



59.000 



A2 



A2 



A2 



At 



Shear 
lb. 



6,200 

5,900 
5,900 

5,900 
6,200 



4.08*8*20=653 



^ 



Joinf 

Coe fficien t 



Columns 



El 



E2 



E3 



[-4 



Ft 



F2 



9,300 



18, 500 



same 



32 1.26 



85 



SO 



82 1. 26 



81 



63 



6.30 



CI 



C2 



CI 



C4 



C5 



C6 



St .63 



32 1.26 



6 *l.5+0+8+8 :< * 9 



C3 .69 



32 1.76 



51 .63 



same 



same 



same 



46,500 



92,500 



93, 000 



92,500 



asjoint 



asjoint 



asjoint 



asjoint 



52 



52 



82 



31 



9.280 
3.250 
9.250 
9,250 
9.280 



F3 



F4 



630*7*20-882 



same 



same 



to, too 



10. too 



same 



same 



asjoint 



asjoint 



50,500 



81 



82 



I Of. 000 




50,500 



asjoint 



asjoint 




101. 000 



82 



61 



9.280 
9,680 


9,680 
9,280 



Of 



02 



63 



64 



HI 



H2 



H3 



H4 



61 



82 



81 



81 



81 



81 



31 



81 



82 



81 



Shear 
lb. 



.63 



1.26 



32 1.26 



.63 



318 



.S3 



52 1.26 



82 L 26 



same 



same 



same 



same 



Moment 
ft fb. 



as joinf 



Beams 



Moment 
ft. fb. 



Shear 
lb. 



Bl 



9SJC/nf[ 32 



* « 



as join 



isjoint 



52 



81 



9.280 
9,750 
9.280 



3.18*4*20: 302 



same 



same 



same 



63 



3.18 



.63 



32 1 26 



82 f.26 



.63 



3.18 



.63 



1.26 



82 f.26 



63 



3.18 



same 



asjoint 



asjoint 



asjoint 



asjoint 



81 



82 



32 



31 



9,780 
9,250 
9.280 



3.18*3*20-221 



same 



same 



same 



asjoint] 8 1 



as join A 82 



asjoin 



82 



same 



asjoinh 81 



9,280 
9.750 
9,280 



3.78*2*20-151 



same 



same 



same 



same 



asjoint 



asjoint 



asjoint 



81 



82 



82 



asjoinh 8 



/ 



9,780 



- 9.750 



9,780 



3.18x1*70' 16 



5.16 



516*6*20=620 



Jl 



01 



02 



03 



04 



OS 



06 



81 



.63 



same 



asjoint 



82 1.26 



same 



82 1.76 



same 



**um«*>M 



15. 900 



A 2 .80 



same 



At 



44 



same 



asjoint 



as joint" 



19, 500 



asjoint 



asjoint 



Bl 



82 



82 

95,000 

64.000 



A2 



At 



9,280 
9,250 
9,380 

6,150 
6,700 



J2 



5.47 



5.4 1*5* 20-- 54 7 



- 



J3 



J4 



fA/0.44) 1. 33 



19.500 



(A2- 0.80) 2.40 



35, 300 



(A2 ■0.80)2.40 



same 



(Af: 0.44) 1.33 



same 



91.500 



176.500 



asjoint 



195,00 



3 



116.500 



J2 



asjoinn J I 



16,580 

17,650 
18.560 



(2.48)7.46 7.46*0*20'00 



Sum of alt joint coefficients -43. 59 

Moment of joint coefficient! with respect to bent J 3458 

Eccentricity if joint coefficients; 600- Jj^| '0.6 ft 
Factor for joint coefficients = *ffi$< 14, 700 



Tabl<- i. Distribution of 



Pressure to Frames Extending East- West 



:u 



Multiplying each joint coefficient by 14,700 gives the column shears 
recorded in the third column. Column moments are computed as the column 
shear times h/2, or 5 ft. At each joint, the sum of the column moments 
equals the sum of the beam moments and will be distributed to the beams 
in proportion to their iC-values. Beam shear is computed as the sum of 
the end moments in the beam divided by the bay length, 20 ft. 

Attention is called to joint D4, where the sum of the column moments, 
2 X 79,500 = 159,000 ft.lb., is distributed to the adjacent beams in pro- 
portion to their K- values as follows : 



1.5 



1.5 + 1.0 



X 159 3 000 = 95,000 ft.lb. to the interior beam, and 



1.0 



15 Q X 159,000 - 64,000 ft.lb. to the spandrel beam. 

Table 5 contains a summary of a study made by combining the floor 
framing in Fig. 17 with columns of varying stiffness. The original X-values, 
4 for wall columns and 8 for interior columns, were reduced to 2 and 4 
and also to 1 and 2, the floor framing remaining unchanged. For these three 
designs, the percentage of wind pressure carried by the individual bents 
was computed, the translation at all joints in a floor being equal. The 
results are recorded in Table 5. 



Bent 



A 
B 
C 

O 

E 
F 

G 
H 
J. 



X-values of Columns 



T and 2 



9.0 

14.4 
12.4 

12.6 
8.7 
8.7 



.7 
.7 
16 8 



2 and 4 



9.2 
14.4 
12.0 

12.6 
8.7 
8.7 

8.7 

8 7 

17.0 



4 and 8 



9.4 
14.4 
11.8 

12.5 
8.7 
8.7 

8.7 

8.7 

17.1 



Table 5 



The percentage of wind pressure carried by each bent is surprisingly 
uniform despite the variation in the i£-values of the columns. This uniform- 
ity will greatly reduce the analytical work required for a group of typ- 
ical floors. 



17. Eccentric Wind Pressure 

If the original if -values for bent J as given in parentheses in Table 4 
are maintained, the floor in addition to its translation due to wind pressure 
will be rotated by a torsional moment equal to the wind pressure multiplied 
by the eccentricity of 9.5 ft. (see Fig. 17), or 

M m 640,000 X 9.5 = 6,080,000 ft.lb. 



35 



The columns resist this torsion and the column shears computed in Table 4 
should be corrected accordingly. 

Mr. Albert Smith, in his paper on "Wind Bracing,"* uses the following 
method of correction. He lets the torsional moment be resisted by shear 
at all the joints of all bents in both directions and determines the shear 
correction, "Tv, in each bent, X, in accordance with the formula 



7v = 



M 



in which 



Ix + Ir 



vxx, 



(11) 



X = Bents A to J, 

r.v = Relative shear in bents X (recorded in Tables 4 and 6), 
x = Distance from bent X to center of gravity of vx (recorded in 
Table 6), 

Ix = v x x 2 

Y, v y , y and I Y = similar values for Bents 1, 3, 4, and 6. 

Equation (11) will, for illustration, be applied to the condition in Problem 
9, using the original Jv-values in bent J. 

Refer to Problem 9 for values of vx and the position of the center of 
gravity. The calculations leading to vy and y are omitted, but may be 
duplicated from data in Problem 9. The values of 7. v and 1 Y in equation 
(11) are computed in Table 6.** 



Bent 


r. v X *» = I X 


Bent 


Ty X V 2 - ly 


A 
B 

c 

D 

E 
F 

G 

H 
J 


4.08X70,5 2 = 20,300 

6.30X50.5*= 16.100 

! 5.16X30.5-'= 4,800 

5.47X10.5 2 - 600 
I 3.78 X 9.5*= 300 

3.78X29.5 2 = 3,300 

3.78X49.5 2 = 9,300 
3.78X69.5 2 = 18,200 
2.48X89.52= 19,900 


1 
3 
4 

6 


6.48X41.32= 11,100 
I 2.64X1 -3 s = 00 

6.67X18.72= 2,300 

2.48X58.72= 8,500 




ly = 21,900 
I x = 92,800 


I x + ly - 114,700 




/ i - 92,800 






Tal 


»lc6 





After having determined Ix 4- ly, the regular procedure would be to 
compute r.v from equation (11) and then cornet the shear in each column 
but the following short cut is more convenient. Compute for each bent, 

-< • Table 7, the value of 



F = 



640,000 , 640,000 X 9.5 



»s.61 



+ 



114,700 



O) = 16,600 + 53.06c). 



in which :;8.61 is the sum of all the joint coefficients in Table 4, using 
•r the bent J. The product.- f F and the joint coefficients in Table l 
equal the final shears including correction for i entricity. 

*.-• i Refcrei ;. 

1.3. 4 and I oly an in led on the right side of Cable 6. 14 in utumnl th< a ihat t» 

J' in' lb-South in Fie 17bav«- itiffneae ii ting wind | K nine 

'•• i<> w a in the North-J- .ill -i. make allowance !• m«l 

Iculatetl MlTii <Mor a* discuss* <J in - tion 12 U uw* m w\ h th 

iiian a panel f> ich i olun i 



',<> 



Bent 



A 
B 
C 

D 

E 
F 

G 
H 
J 



M 



Ix+I* 



-(*) 



53.0 X( -70.5) 
53.0 X( -50.5) 
53.0 X( -30.5) 

53.0 X( -10.5) 
53.0 X(+ 9.5) 
53.0 X( +29.5) 

53.0X(+49.5) 
53.0 X( +69.5) 
53.0XC+89.5) 



-3,700 
-2,700 
-1,600 

- 600 
+ 500 
+1,600 

+2,600 
+3,700 
+4,700 



+ 



F = 16,600 
M 



Ix + I 



(x) 



14,700 



12,900 
13,900 
1 5,000 

1 6,000 
1 7,1 00 
1 8,200 

19,200 
20,300 
21 ,300 



0.88 
0.95 
1.02 

1.09 
1.16 
1.24 

1.31 
1.38 



Table 7 

The last column in Table 7 gives ratios of F/14,700, which is the ratio 
of column shears with and without eccentricity, see also Table 4. When the 
eccentricity is 9.5 ft., the shear varies considerably from bent to bent, the 
shear in bent H f for example, being 38 per cent greater than it is when 
no eccentricity exists. It is seen that changing the stiffness of a bent may 
greatly affect the distribution of wind pressure; and for this reason, values 
of K should include allowance for effect of haunching as discussed in 
Section 14. 

18, Accuracy of Approximate Analysis 

In order to study the accuracy of the results obtained by an analysis 
as applied in Section 16, two frames were analyzed for wind pressure. The 
first frame is one which Professors Wilson and Maney have analyzed by 
the slope-deflection method;* the second frame has been analyzed by Mr. 
John E. Goldberg** by a method of converging approximations. The results 
of the analyses are given in Fig. 18 for the third floor only; they are typical 
for most of the floors in the bent. 

The moment values in parentheses in Fig. 18 are taken from the sources 
mentioned. Moments obtained by the procedure used in this text are under- 
scored, and typical computations are as follows. 

The joint coefficients in Fig. 18(a) are 

21 4 
at exterior columns: 35.4 X tt— — : — ^— — — r^— : = 8.20 



21.4 + 35.4 + 35.6 

at interior columns: 35.5 X ' ' 

21.4 + 29.2 + 35.5 -f- Sb.b 

Sum of coefficients for four columns: 



= 14.75 



45.90 



The column moments, in in. lb., are obtained by multiplying the joint 
coefficient by 

6,690 X (7 X 12) 



45.90 

7,140 X (8 X 12) 

45.90 



= 12,240 for columns above 3rd floor 



= 14,930 for columns below 3rd floor, 



in which the values of 6,690 and 7,140 are the wind shears given in Fig. IS 



*See Reference 21. 
**See Reference 34. 



37 



The moments in beams are obtained by distribution of the sum of the 
column moments; thus, the end moment in the center span is determined as 



(181,000 + 220,000) X 



29.2 



29.2 + 21.4 



= 231,000 in.lb. 



6.690 



5^> 





(90.1) 
3 r -4 Floor* 222 



(203.0) 




2/4 



12? 



* v42 0) 




032.0) 

no I 23/ 

(184.0) I (226.0) 

220 X 

(d) 




(226.5) 




6.69 




* ** 



18-0 




t **r 



22 : 



115 



(109) 
3 c? 'floor I 256 



(226) 




% 



206 

(211) 
195 I 266 




(206) I (252) 
253 



'9.2 



(258) 




(b) 
Fig. 18 



Figures circled in are values of j 
Other figures are moments at joints 
due to the wind pressure indicated. 



figures underscored are obtained 

by approximate procedure. 
Fig ure s in paren thes is a re 
computed by: 
fa) Wi I son-Maney(Reference2l) 
(b)Ooldberg (Referenced) 

18-0 




(172) 
262 I 1/5 



(250) I (122) 

207 

(210) 






The frame in Fig. 18(b) is the same as that in Fig. 18(a), except that 
the stiffness of Columns C and D and the intermediate beam has been 
reduced as shown. The procedure in determining the moments underscored 
in (b) is the same as that used for moments in (a).* 

According to studies made by Mr. Albert Smith, moment values such 
as those underscored in Fig. 18 may be corrected by the following procedure : 

(1) compute a correction equal to one-fourth of the difference between the 
larger and the smaller moment at the ends of a member (beam or column), 

(2) deduct the correction from the larger moment, and, (3) add the cor- 
rection to the smaller moment. Consider, for illustration, the end span in 
Fig. 18 (a) and compute: (1) the correction % (222 - 170) - 13, (2) the 
corrected moment to the left 222 - 13 = 209, (3) the corrected moment 
to the right 170 + 13 = 183. By such corrections, some allowance may 
be made for the fact that the point of contraflexure is not at the midpoint 
as assumed in the derivation. 



*The moments in parentheses in Fig. 18(b) do not satisfy the requirement that their sum equals tero 
at each joint. At column A. for instance, the sum of column moments, 109 + 133 = 242, should equal 
the beam moment, but the beam moment is given as 226. These moment values are approximate. 



38 



19. Moment Deflection and the Ideal Layout 

The beam shears in Table 4 are practically equal in all spans of a regular 
bent. In such bents, the sum of shears is small at the interior columns, 
and these columns will receive little or no direct load from wind pressure. 
The direct load in the exterior columns equals the shear in the end span 
and is about 9,300 lb. in bents, B, C and E to H, the direct load being 
tension in the windward and compression in the leeward column. The result 
is a differential column extension and a warping of the floor, which causes 
a "secondary" distribution of shears and moments. 

It is possible to arrange a structural layout so as to minimize the effect 
of differential column extension and thus approach an "ideal layout" which 
may be accurately analyzed. In the ideal layout there is (1) no wind pres- 
sure eccentricity and (2) no warping of the floors. The building frame 
should be laid out with symmetry in both directions; or, at least, the wind 
pressure component should go through the center of gravity of the joint 
coefficients, a case which was illustrated in Problem 9. To avoid warping 
the floors, the interior bays should be designed to carry much larger shears 
than the end bays. This may be accomplished by making the coefficients 
at exterior joints sufficiently small compared with the coefficients at 
interior joints. 

To illustrate the procedure in minimizing the warping of floors, A'-values 
for bent B in Fig. 17 and Table 4 will be adjusted as follows and the wind 
pressure distributed accordingly, the calculations being recorded in Table 8. 



Span 


BI-B2, B5-B6 


B2-B3, B4-B5 


B3-B4 


X-value 


1.0 ' 


1.5 


1.7 



Bent B is assumed to receive the same total wind pressure, 92,600 lb., as 
in Table 4; and shears, moments and direct column loads are computed 
in Table 8 by the same procedure as used in Table 4. 



Joint 


Joint Coefficient 


| Columns 


Beams 


Direct Column Load 


Shear 


Moment 


Moment 


Shear 


Adjusted 


Original 


1 m 


0+1 
/\ v — n AAA 


7,200 


36.000 


72,000 


7,100 


+7,100 


+9.300 


0+1 +4+4 


B2 


8X 1+^+8+8 = 1 " 080 


17,500 


87,500 


70,000 
105,000 


+3,220 





1 0,320 


B3 


HV 1 - 5+1 - 7 1333 


21.600 


108,000 


101,300 
114,700 


+1,150 





U X 1 .5+1 .7+8+8 _1 "" J 


11,470 


£4 


1.333 


21,600 


108,000 


114.700 
101,300 


-1,150 





10,320 


■ 

B5 


1.080 


1 7,500 


87,500 


105.000 
70,000 


-3,220 





7,100 


56 


0.444 


7,200 


36,000 


72,000 


-7,100 


-9,300 



Tablt- 8 



M) 



The direct column loads determined for the original and also for the 
adjusted K- values (see Table 8) are plotted in Fig. 19. The curve marked 
"ideal" in Fig, 19 indicates the magnitude of direct column loads that will 
cause no warping of the floor and, therefore, no secondary distribution of 
moments and shears. It is seen from the curves in Fig. 19 that a much better 
condition for accurate wind pressure analysis is created by selecting pro- 
portions of beams in accordance with the K- values suggested. From 
inspection of the joint coefficients in Table 8, it is seen that the adop- 
tion of new column sections may be 
similarly effective. It is especially 
advisable to make the wall columns, 
Bl and B6, rectangular with the 
smaller dimension in the East-West 
direction. 

The effect of warping of floors 
becomes increasingly serious the 
shorter the outer span is compared 
with interior spans. In such cases, 
warping may be minimized and the 
shear reduced in end beams by 

judicious Selection Of Shallow OUter ^^ Mi^tes are direct column load due ro»ind. 

beams and narrow exterior columns. Fig. 19 




H.(i>00 
1 1 00 

9.300 



20. Illustrative Problem No. 10; Vertical Load and Wind 
Pressure Combined 

The problem is in two parts: (A) Analysis of bent G in Fig. 17 for vertical 
loads, and (B) determination of maximum moments and shears due to 
vertical loads and wind pressure combined. 

(A) Analysis for Vertical Loads. Bent G in Fig. 17 has three spans, each 
of which has a length of 20 ft. from center to center of the columns; the 
width of the bay carried by bent G is 20 ft. The successive steps in the 
analysis will be the determination of 

(a) Loads. 

(b) Values of K or n. 

(c) Values of joint coefficients and moment coefficients. 

(d) Fixed end moments. 

(e) Moments at ccnterlines of columns. 

(f) Moment reduction for effect of width of columns. 

(g) Shear. 

(h) Maximum values required for design. 
(a) The loads in lb. per lin. ft, of the beams will be taken as 



Dead Load (D.L.) 

Live Load (L.L.) 
Total Load (T.L.) 



75 X 20 + 300 
100 X 20 = 



1800 
2000 
3800 



hi 



(b) The A'- values* will be the same as those used in the wind analysis 
of Bent G (see Problem 9, Section 16) : 



Wall Columns: K 
Interior Columns: K 
Interior Beams: A' 



= 4 
= 8 
= 1.5 



(c) Values of joint coefficients and moment coefficients based upon the 
A-values in (b) are recorded below, see summary in Section 10 for equations 
and references: 



Joint Mark 
See Fig. 20 


~TT =n 


Beam Moments 


1 

Column Moments 


Value of Q 


At Ext CoL 


Exterior 


Interior 


1 


4n 
4n+1 


2n 

4n+1 


3 71 

6n+2 


4n+2 


G\ 
G1 


4 

T.5 

8 
1.5 


0.079 
0.043 


0.92 

• ■ m m 


0.4$ 


m a ■ a 

0.47 



(d) Fixed end moments, in ft. lb., using 
in Fig. 1, are** 

D.L.: -Via X 1800 X 202 = 
L.L.: -Via X 2000 X 20 2 = 
T.L.: -V I2 X 3800 X 20 2 = 



the coefficient in case 5 



-60,000 
-66,700 
-126,700 



(e) Moments at centerlines of columns, computed according to the pro- 
cedures in Sections 2 and 7, are numbered below. Corresponding numbers 
are written on Fig. 20 to indicate the point where the moment is taken 
and the loading arrangement which produces it. 




03 



04 01 



02 



03 



64 01 



62 



63 



64 



* 



8 
















9 10 








& 




Fig. 20 













1. 



2. 



3. 



4 



Maximum end moment in G\ —G2 at Gl : 
M x = 0.92 X (-126,700) = -116,500 

Maximum column moment at Gl : 
Af 2 = 0.46 X 126,700 = 58,300 

Maximum column moment at G2: 
Af 3 = 0.47 X 66,700 = 32,00 

Maximum end moment in G1—G2 at (72: 
M 4 = -126,700 - 1 X 0.079 X 126,700 



136,800 



♦When cross-sections are not known, select values for in accordance with the classification suggestc I 
in Section 3. If the cross-sections are known, values of I — and iC— may be determined by using equations 

(7), (8) or (9) in Section 12. 
**For numerical example involving unequal spans, see Problem 7. 



41 



5. Ma ouni ctnt< r in«>m<i)t Gl — < * 

• M:-l 0+2X0.079X1 0+1X0.043X (-66,700) « — 109,000 

at < -1 jki- 1XQ( >X\ 7Q0-2X <»"1 -6< ,:<hi «_- 1 ;u,100 
age: - 120,300 

3/ 5 - + X 8800 X . - 120,300 - ■ 

6. Mai m center d n< in G2 * 

d t 

-12 00 + (2X0.079 IXOJ X 06,700 - - J J 1,400 
M 6 - + x 3800 X KP - 121,100 - +68,600 

l jim I a: G] to t>< used for maximum shear at 

U, - i 700 + 2 x 0.078 X 128 - 106,600 

(f; M i effe of width oi oolunu - follom, ' 

iiiiiu *. hmoii l hum' 1 ' < i iii inil< on from Gl i < < 

l HI (MM) s linn 1) 

-1/, 1 I f Iaj X 3SiMiit x : ■ — M »o 

lf< - • ) - ';< X 3£ Kl X 2 - - I" Ml 

aft -4 • - ' X 88,000 X : - H 89, mi 

U„ - + <;v ,i! - i , x 000 x - 4 hi 

< at tin- far In .lu it, tl unlll. f Wl t) if 

' • ' : • ' ' - < "luptlted I f<ll|c>W>.*** 1 ll«' 1 1 1 1 f 1 1 1 •< T- iK'loW U f«TM|l| 

»«- I- ii I 10 a it li load < <i i . i ma iun la. 

4 d Gl 

1 « S00 X (KM) 



|. M I iilM -ii at 

1-380,.' ! " l: ,,.„,,, 

I < ...» 

M «hm 3 at t 



J/y - 121 I / 

U H , » -1 X 66,700 

k \i r roo - - 



- 

i I 



• » it 

i/. mm »• - 



I 







(h) Maximum* values required for design are as follows: 

Moment in beams: M lt M i} M 5 , M & . 

Shear in beams: F 4 , V$, Fg. 

Direct compression in beams is disregarded. 

Moment in columns: M 2 , A/ 3 . 

Shear in columns is disregarded. 

Direct compression in columns taken from the customary 
computations of column loads needs adjustment because 
the frame action transfers some column load from Gl to 
(?2. An increase in column load of 5 per cent (see F4 = 
33,300 + 1,700) at (72 is conservative, but no decrease 
in column load at Gl is warranted in this example. 

(B) Maximum moments and shears due to vertical load and wind pressure 
combined. The frame is the same as that analyzed in (A) for vertical load 
and in Problem 9, Section 16, for wind pressure. 

Using the centerline moments determined in Table 4, the beam moment 
in ft.lb. at the face of all the columns in Bent G is 

m w = * i ::■::: ) x --—— = * si, 000 

The maximum moments, Mi, M4 and M 5 , in the end span computed 
in (A) are plotted in Figs. 21(a) and (b) and connected by a parabola, which 
will be considered the curve of maximum negative and maximum positive 
moments due to vertical loading.** Beam moments due to wind pressure 
are indicated by the lines connecting ±8 1,000 with =f 81,000 in Fig. 21(a) 
and (b). The combined moments are measured between the inclined line 
and the parabola. 

The maximum positive and maximum negative moments taken from 
Figs. 21(a) and (b) are plotted in Fig. 21(c). It is seen that wind pressure 
added to vertical load increases the positive moments and also the negative 
moments at both ends of beam GI-G2; the length in which there may be 
tension in the bottom or tension in the top of the beam is also increased. 

The maximum shears, F 4 and V&, are plotted at Gl and (72 in Fig. 22(a) 
and (b) and connected with a straight dotted line which will be considered 
the curve of maximum shear due to vertical loading.** 

The shear in the beams at the face of the columns due to wind pressure 
is constant and equals 

F. = 81fi0 ° + 81 ' 00 ° = 9,300 

17.5 

The solid, inclined lines in Figs. 22(a) and (b), drawn at a constant distance 
of 9,300 from the dotted lines, represent the combined shear. The maximum 
shears taken from Figs. 22(a) and (b) are plotted in Fig. 22(c). It is seen 
that the shear may be increased over the entire length of the clear span. 

♦For discussion of minimum values of moments and their significance in detailing bar reinforcement 
in frames with unequal spans, see page 22 in Section 9. 

**In the intervals between the points plotted, moments and shears may be numerically larger than those 
indicated. The discrepancies are frequently small and will be disregarded in this ease. 

43 



01 



(3) 

Direction of Wind 



*59.IOOtMr) 



02 

-61.000 



*32, 100 



(a) 
Direction of Wind 



mooo 

-64.800 
(M t ) 




(b) 
Direction of Wind 




59. 100 (M s ) 




fy D irection of Wind 



02 



CO 

Max. combined moments 

*59.IJfO(M<) 





01 M 

+4l,40d^Max. combined shear 



w 

t -35, 000 

(y 4 ) 

02 




Fig. 21 



Fij£. 22 



Tin- principles embodied in a comprehensive analysis for wind pi sm 

have been presented, discussed and illustrated in the foregoing rliuri; 

For the Bake of simplicity, the examples have been prep d for compara- 
tively simple structural layout. For a frame with highly intricate wind 
■lion, Mr. Albert Smith has published a detailed solution for one floor, 
Hcferen* 56 in the bibliography. In the solution on pages 919 and 

920 — it should be observed — the calculations for joists and Mush beams 

might have been lumped in an approximate total without ther v intro- 
ducing appreciable error in the final moments in the important I in 
More than one-half of the calculations given might then have been omitted 



ti 



APPENDIX A: DERIVATION OF FORMULAS 



VERTICAL LOAD 



21. Angle Changes Determined by Moment Area 
Principle 

A prismatic member connecting two joints, A and B, will deflect when 
loaded as illustrated in Fig. 23. The angles between the line connecting 
A and B and the tangents to the elastic curve at A and B will be denoted 
as Q A at A and 6# at B. The relationship between these angles and the 
loading may be established by the moment area principle.* 

Let ACDB be the moment curve due to the loads, P, placed on the 
simply supported beam AB. Apply the cross-hatched area — the moment 
:irea — as vertical load on a beam similar to AB and determine the end 
shears, V& and V&, produced by this Loading. According fco the moment 
area principle,** the angle changes in Fig. 2'A are 



<>i = ,!, X I .t.and B B = ,| T X V B 



in which 



EI 



E = modulus of elasficitv and 



EI 



(12) 



/ = moment of inertia of the cross section of .1 B, 



A 



Pc 



Pr 



3 





Fig. 2.'{ 



Equation (12) expresses that "the angle changes at .1 and B in Fig. 23 
equal (1/EI) times the end shears due to the moment area applied as 
load on beam AB." This is a form of the moment area principle, from 
which the procedure followed in frame analysis may be developed. 

It is customary to define signs for end shear due to vertical loading as 
positive at the left and negative at the right support, and the same con- 
vention applies to angl< ■ hanges. Further discussion of sign conventions 
is given in Section 27. 

Problem 11. (a) A simply supported beam, ^.5„has a load, w } uniformly 
distributed over the entire span length, I. Determine the angle ehang< 
at A and B. 






*The premutation of the moment area principle ilates bark to Otto Mohr who, in I vbUahi 

" method of computing deflections as if they were bending momenta The lat- (' E Greene at ti 

University of Michigan, about 1873, discovpred a related principle of moment! m. H. Mull>T-Breslau. 

1885, gave Mohr'a principle a general formulation. 

♦♦Derivation of the principle is given in many textbooks =>uch aa Referen I (p 10) and Reference 1 l 
(p. 210). 



15 



ID I 

The moment curve is a parabola with a maximum ordinate of — . Load 



8 



beam A B with the area under the parabola which equals % X 
determine the end shears : 



T 



and 



V A = -Y B = MX y 3 x 



wl s wl z 



b 



24 



The angle changes, according to the moment area principle, equal 



o A = -e B = 



24JET 



* • * • 



(13) 



(b) A concentrated load, P, is placed on beam AB at a distance of al 
from A and bl from B; show that the angle changes due to the load, P, are 



A = ~6EI ( ' an = 



PabP 
6EI 



(1 + <0 



(14) 



22. Angle Changes Due to End Moments 

The loads on beam AB in Fig. 24 are two end moments, M AB and M BAr 
which deflect the beam as indicated. Let the angles between the line AB 
the tangents to the elastic curve at A and B be denoted as Q A at A 
and Q B at B. The relationship between these angles and the moments that 
produce them may be established by the moment area principle. 






Fig. 

The moment curve is the straight line connecting points with ordinate* 
M AB at A and M BA at B. Load beam AB with the cross-hatched moment 
area ACDB in Fig. 24, divide the trapezoid into two triangles as shown, 
and determine the end shears: 



v, = 



.' 



i; = 



B 



(HMab xi)xy 3 + (y 2 M BA x o x H = 

HMab X / + HMba x i 
-('AMab X I) x M 



^ ab X I 



Wba X J) x h 

BA X I 



. (15) 



According to the moment area principli xpressed in equation (12), 
the angle changes 8' in Fig. 24. derived from equations (15) by selling 

1 

0' = — X V\ are 

hi 



Q'a = J^ X (23/^ + V flA ) 



n, = 






X (2Jl/« x + 3f iB ) 



( 1 ' 



16 



Equations (16) give angle changes in terms of end moments. This relation- 
ship is important in frame analysis. 



23. End Moments Required for Fixity 

To determine the fixed end moments, Ml B and M£ At due to a load, P, 
consider first the beam AB simply supported as in Fig. 23, the tangents 
being rotated through angles expressed by equations (12). Then apply, as 
in Fig. 24, two end moments of such magnitude that the tangents at A 
and B are brought back to their original position. The latter step is equiv- 
alent to fixing the two ends of beam AB, and the end moments, M% B 
and M BA , equal the moments at the supports of a beam with fixed ends. 
The procedure is expressed algebraically by the equations B A + 9^ = 
and 05 + Qb = 0, the 9- values being taken from equations (12) and (16). 
The equations become 

V A - -J(2M & + Mg A ) 

V B = +|(2M& + M F AB ) 

6 



F 



Solving for M* gives 



M F AB = -|(27j + V B ) 

2 > (17) 

MSa = +~(2V B + V A ), 



s 



in which V A and V B are the end shears due to the moment area for load P 
applied as vertical load on the simply supported beam AB. Since V A is 
positive and V B is negative, the fixed end moments are negative quantities. 
Further discussion of sign conventions is given in Section 27. 

Equations (17) are used for determination of fixed end moments, such 
as those presented in Fig. 1. 

Problem 12. (a) A beam AB has a load, w, uniformly distributed over 
the entire span length, I. Determine the moments, M AB and M BA required 
to fix the ends of AB. 

Inserting in equations (17) the values 

V A = -V B = U ^ 
A B 24 

as determined in Problem 11 gives 

M F AB = M F BA = -^ (18) 

(b) A concentrated load, P, is placed on a beam AB at a distance of al 
from A and bl from B; show that 

Mi B = -ab 2 (Pl), and Mg A = -a?b(Pl) (19) 

Equations (18) and (19) are used frequently in frame analysis; they 
should be memorized for convenience. 

47 



24. Moments as a Function of End Rotation and Fixed 
End Moments 

If joints A and B in Fig. 25 are locked artificially before the loads P 
are applied, fixed end moments — M% B and M$ A — will be created which 
tend to rotate the joints. When the artificial restraint is removed, the joints 
will rotate through angles, Q A and B in Fig. 26, until equilibrium is 




P. 



M 



8A 



; 




-)( 



Joints A and 3 locked 



3 




Joints A and 3 released** 




Fig. 25 



Fig. 26 



attained. New end moments, M' AB and M BA , are thereby induced in AB 
in addition to the moments M F that already existed. The final moments 
are then 

M A b = Ml B + Ma B} and M BA = M| A + M BA - : - (20) 

The values of the moments induced by the angle changes may be deter- 
mined from equations (16) by introducing M' as the moments induced by 
the angle changes, 0: 

2M' A b + M BA = GE X - X e A 

M' AB + 2M BA = -6E xjxOi,. 
Solving for M AB and M BA gives 



Mar = 



itm* = 



2E X ~ X (26 A + Q B ) ) 
/ . . „ , 






(21) 



which inserted in (20), - being denoted as K, gives the final moments 

Mab = M f ab + 2 X (2EKB A ) + 1 X (2EKQ B )\ 

M BA = Mg A - 2 X (2EKQ B ) - 1 X (2EKQ A )f . . (22) 

Equations (22) are the Sloj -Deflection Equations; they express 
moments in frames in terms of M' and joint rotation. Joint translation i 
disregarded in equations (22) in accordance with common practice 

Equations (22) express thai an end moment, M AB} in a beam .1 H whh-h 
is part of a frame, may be determined as the algebraic sum of I jiree terms: 

(1) fixed end moment I A: M ABi 

(2) moment induced by joint rotational t h<- near end, A : 2(27:' A' o., ), and 

(3) moment induced by joint rotation at the far end, B: \(2EKO B ). 

Fixed end moments may be determined when loads and spans ao trivr q 
The vain of EK& used in (22) will b< giv< □ furl her stud- 



18 



25. Joint Rotation a Function of Stiffness 

Fig. 27 shows four members which are part of a frame and intersect at 
joint A. Let joint A be rotated through an angle, 9^, by a moment, U Af 
applied at A. At the far ends of the members, four different conditions 
are assumed for illustration: joint B is fixed (Q B = 0); at joint C the rota- 
tion is —e A} at D it is +9a, and at F it is fQ A , the value of / depending 

upon the conditions at F. The y -values will be denoted as K ABi K ACt K AD 

and K AF , which represent the stiffness of the various members. 

The joint rotations in Fig. 27 have induced end moments which may 
be expressed by use of equations (21). Since joint A is in equilibrium, the 
sum of the end moments at A must equal U A . This requirement is expressed 
in the summation that follows:* 



M AB = +2EK AB X (2G A + (0)) 
M AC = +2EK AC X (29^ + (-9 A )) 
M AD = -2EK AD X (29 x + (+0J) 
M AF = -2EK AF X (29 A + (fO A )) 



= +[2EK AB (2e A )} 
= +[2EK AC (B A )) 
= -{2EK AD (3 G A )} 
= -[2EK AP (2 + f)9 A ] 



2M AX = U A = 2Ee A X [2K AB + K AC + 3K AD + K AF {2 -f /)]. 



M A b 



%**>% 



M*£25JWA 



F 




e B *o 

(f* 0) 




M AC =K AC (2E0 A ) 



2BQ A - 



' A UK AX *(2*f x ] 
in which x is 3. C DorF 

M A F*K„f*f)(2£9 A ) 




(d) \ M AD 

Moments induced by U A 



f2FQ A ) 




/[& =+& A 



Fig. 27 

The four moments induced by U A are shown diagrammatically in Fig. 
27(a). Since all four moments tend to rotate joint A in the same direction, 
it is the sum of the quantities within the four brackets above the summation 
line that equals U A . 

*For discussion of signs used in the derivation of equation (23), see Section 27. 



49 






The equation XM A x = U A gives 

U 

2E ° A m 2K AB + K AC + ZKad + K AF (2+/) ' ' ' (23) 

Equation (23) expresses that 2EQ A equals U A (unbalanced moment due 
to the fixed end moments at A, see definition in Section 27), divided by 
the sum of the products K X (2 + /) written for each adjacent member. 
By comparing equation (23) and Fig. 27, it is seen that 

2 + / = 2, when angle change at far end equals : / = 
2 + / = 1, when angle change at far end equals — Q A : / = —1 
2 +/ = 3, when angle change at far end equals +Q A : f — +1 

Equation (23), written in its general form, becomes 

^'- zgJW (24) 

26. Derivation of Formulas for Joint Coefficient, Q 

A quantity denoted as Q is introduced and defined in Fig. 2. It will be 
seen by comparing Fig. 2 with equations (22) that the value of Q A is 

An expression for the right side of the equation is obtained by multiplying 
with —y2- in equation (24), which gives 



Qa = 



2EK AB Q A K A b /ggx 

U A X[K AX (2+f x )} 



Therefore, Q A is a function of 

K A x'- the I/l- values of the members intersecting at joint A, and 

fx : the degree or type of restraint at the far end of the members. 

An approximation is now introduced by inserting in equation (25) : 

2 + fx ~ 2 for all columns, 

2 -\- fx = 2 for beam adjacent to exterior joints. 

2 + fx — 1 for beams adjacent to interior joints. 

If, in addition, the stiffness of beams adjacent to any one joint is set 
equal to K and the stiffness of columns at the same joint equal to nK, 
equation (25) may be written in the simple form : 

Q = — — .** (26) 

* 2(2tl&) + 2K 4n + 2 

Either equation (25) or (26) may be used to determine Q. The procedure 
in Steps (1) to (6) in Section 2 would give mathematically exact results 

*A quite accurate and very quick analysis may be made by designers who are accustomed to judging 
the relative magnitude of joint rotations. The procedure is simply to estimate the value of /for the far ends, 
determine joint rotations from (241, and compute the final moments from equations (22). 

Professor Maney has developed an approximate method involving the following steps: (1) at alternate 
jointB set/i = and calculate 9 from equation (24) ; (2) using these 0-values, calculate at the intermediate 
joints; (3) re-calculate G at joints in firstgroup; (4) repeat steps (2) and (3) if necessary: (5) calculate moments 
by equations (22) . This method was presented by John E. Goldberg, see Reference 9. A similar procedure 
has been presented in a manuscript by Raymon C. Buell. 

**For roofs, the equation becomes Q - = — iTnt based on assumptions similar to those made for floors. 

50 



if / in (25) were known; it becomes an approximate method* because / 
is estimated or approximated as in (26) . 

The value of /appears always together with K in the product of K(2 + /) . 
The i£-values are not known beforehand, but must be estimated before a 
frame can be analyzed; it is then reasonable to estimate / also and to 
use the simplified equation (26). When reviewing a frame with given pro- 
portions, the "physical characteristics of the structure make it impossible 
to determine the actual bending moment to a greater degree of accuracy 
than =*=5 per cent."** Therefore, even when proportions are given, a rela- 
tively poor estimate of/ is of little consequence. 



27 * Conventions and Use of Signs 

Sign conventions used in current practice and their application to the 
derivations in the foregoing analysis will be discussed. 

A beam, AB, through which an arbitrary vertical section S-S is drawn, 
dividing the beam into two parts { M" and "f?," is shown diagrammatically 
in Fig. 28(a). Remove "B" and add to "A" the internal stresses, C, T 
and V, exerted by "B." The stresses C and T form a couple, the internal 
moment; since "A" is in equilibrium, the internal moment equals the couple 
formed by the loads and reactions on "A," the external moment. Trans- 
posing the procedure by removing "A" and adding in S-S the internal 
stresses exerted upon "B" gives an internal moment which is numerically 
equal to the internal moment on "A" but opposite in direction. 






DOS 

Points of contra flexure 




Fig. 2ft 



Fig. 29 



The internal moments at any section may be indicated by two curved 
arrows as in Fig. 29(a). If the section is taken immediately outside joint 
A in Fig. 29, the arrow which is closer to A indicates the moment with 
which beam AB tends to rotate joint A, and the other arrow indicates 
the moment exerted by the joint upon the beam. 

♦Various analytical procedures have been proposed in which a relative stiffness such as K X (2 + /) is 
introduced. See, for example, Professor L. E. Grinter's discussion of "A Direct Method of Moment Distri- 
bution." Proc. A.S.C.E., March, 1935. page 426. 
♦♦Professor L. E. Grinter in Proc. A.S.C.E., March, 1935. page 428. 



51 



The internal force V, the shear, in section S-S in Fig. 28(a) is defined 
as the algebraic sum of the reaction, R' } and the load, P'\ on the part of 
beam AB to the left of S-S. The common convention is to take the shear 
at the end of the beam as positive at the left support as in Fig. 28(b); 
the end shear is then negative at the right support. Apply this definition 
to the type of end shear treated in equations (12) and (16) ; for a deflection 
as in Fig. 23, becomes positive at A and negative at B, and these signs 
are reversed for a deflection as in Fig. 24. From the definitions of shear, 
it follows that fixed end moments are negative (see discussion of equation 
(17) in Section 23). The sign conventions for shear and moment are inter- 
locked by the moment area principle expressed in equation (12). Since 
fixed end moments for vertical loading are negative and occur where the 
elastic curve is humped, "a moment is positive when it sags the beam;" 
this will be referred to as si^n convention number (1). For columns, the 
ime sign convention may be applied if the sheet is turned to read from 
the right. Positive moments will then create tension in the bottom of beams 
and in the right hand side of columns. 

In dealing with joint rotation, it is convenient to adopt another conven- 
tion, number (2): a moment exerted upon a joint by a beam is positive 
when it tends to rotate the joint in the clockwise direction. * Convention 
(2) would make all the moments at joint A in Fig. 27(a) negative. This 
sign convention is preferred in analysis, the determination of rotation and 
moments. The ultimate purpose of analysis is design, or determination 
of stress ; and for this purpose it is customary to use sign convention (1). 

Fig. 20 shows deflected members in part of a frame, and internal moments 

arc indicated by double arrows. The moment equals zero at the two points 

in AB where the curvature changes from a sag to a hump, the points of 
contraflexure. The moments, positive between and negative outside tin 
points, according to convention (1), may be plotted indicated in Fig. 
29(b). If convention (2) were adopted, the momcnl in Fig. 29(a) would 

become positive at A but negative at B; this would be confusing in plotting 

moment curves. Convention (1) is adopted in this text, although l tain 
derivations may appear less direct; but th< application of formulas and 

procedure has doubtlessly been simplified. 

The numerical value of r A is defined as the algebraic differ* ace between 
the fixed end moments at A. According to equation 2 1), V has the same 

BlgQ as &. This requirement will be fulfilled in Fig. 2 when U at an\ point 

is taken as the algebraic different. U M ! to the left minus M r to the righl 

of the joint," the fixed end moments being w< d with their prop< n- 

If at joint A t for ex mple, M AJi > s numerically larj. than M\,, then 
U A = M', - M is p itiv and o also become- positive; if Mis ] " 
maller than \1 \, . both Ua and B A are negative. 



K'.fereii'i ], page 14 



52 



28. Formulas for Moments in Columns 

Let K c and Kb be the J/Z-ratio for columns and beams in Fig. 8(a) and 
set n = K c /K}). The deflection of the columns in Fig. 8(a) is similar to 
that of member AD in Fig. 27 for which/ = +1; therefore, K c {2 + /) = 
3K C . The beams in Fig. 8(a) deflect as member AC in Fig. 27 for which 
/ = —1, and Kb(2 + /) = Kb. Insert SK C and Kb in the equation in Fig. 
27 which gives the relationship between 6 and U in Fig. 8(a) : 

2EQ = (27) 

2(SK C ) + 2(2Q,) 

The end moment, M, induced by joint rotations as in member AD in 
Fig. 27 is 

M = 3K c (2Ee), 

which combined with equation (27) gives 

M = Kr „ W ' rjr = n Zn n XU (28) 

C 6K C + 2Kb 6n + 2 

At exterior columns, with loading arranged as in Fig. 8(c) to give approx- 
imately maximum column moment, the value of K c (2 + /) equals 3A' C 
as for interior columns; but the shape of the deflected beam in the end 
span is somewhere between the shapes of AB(f = 0) and AC (J = — 1) in 
Fig. 27. Adopting/ = -0.5 gives iQ>(2 +/) = 1.5Kb, and equations (27) 
and (28) become modified as follows : 

2ge - 2(3* C ) +!(!.«») ^ 

and 

M = — ^— X U (30) 

4n + 1 



APPENDIX B: DERIVATION OF FORMULAS- 
WIND PRESSURE 

29. Assumptions and Distribution of Wind Pressure 

Let A f B, C, D and F in Fig. 30 be joints in a bent which is deformed 
by bending due to wind pressure. While investigating the conditions around 
joint A, make the following assumptions:* 

(a) joints F, A and C lie on a straight line 

(b) joints B, A and D lie on a straight line 

(c) the same angle change, B A , exists at F, A and C, 0^ being measured 
from a horizontal line 

(d) the same angle change, G A , exists at B } A and D, A being measured 
from vertical lines. 

*Suggested by Professors Wilson and Maney on page 25 in Reference 21 . 

53 






The assumptions, incorporated in Fig. 30, place the point of contraflexure 
at the midpoints of all members. They also provide for a method of distri- 
bution of wind pressure as derived in the paragraphs which follow. 

The part of the bent shown diagrammatically in Fig. 30 will deflect 
under wind pressure, being the angle change at the joints and R an 
angle representing translation of the joints. 



3 




Moments induced by joint translation 




Fig. 30 

The sum of the moments induced in the members at A by the joint 
translation equals zero because A is in equilibrium and no forces are applied 
between the joints. The induced moments, expressed by equation (21), 
are written below, the signs being in accordance with convention (1) in 
Section 27. Fig. 30(a) shows that the moments in the beams tend to rotate 
A in one direction and the moments in the columns tend to rotate A in 
the opposite direction. It is, therefore, the sum of the four moments within 
the parentheses that equals zero. 



M AC 

M AF 

M AB 
M AD 



+2BK AC X[2e A + { + eA)} 

-2EK AF X[26 A + (+e A )] 
+2EK AB X[-3(R-e A )] 

-2EK AD X[~3(R-e A )] 



+ (6EK AC XQ A ) 
-(6EK AF XS A ) 

+ (6EK AB XQ A - 
- (6EK AD X Q A ■ 



6EK AB X R) 
6EK AD XR) 



2M AX = Q = 6Ee A X2(K AX )-GERX(K AB +K AD ). 



54 



The equation below the summation line reduces to 



= e A X (2K AX ) - R(K AB + K AD ), 



from which 



6. = R X 



K AB + K AD 



ZK 



(31) 



AX 



Combine the equation 



p _ a = 7? v (E—**l ~ ^ AB - Z ^p) = /? v — ^ ~^~ ^ AF 

" Tr 4i 2K A x 




with 

M AB = -$EK AB (R - 9 A ), 



which gives 



^ac + Kaf 



M AB = -§ERXK AB -^-^^ (32) 

&&-AX 

If the shear in column AB is denoted as V A , then 

V A =\XM AB = -^? X K AB (^+J^), . . . (33) 

in which the numerator within the parenthesis is the sum of the 7C-values 
for the beams at A and the denominator is the sum of the i£-values for 

all members at A. 

The value of R, the only unknown in equations (31), (32) and (33), 
need not be determined when R/h is constant for all columns in a story. 
In this case, the relative values, v A , of the shears, V A , taken by each column 
become equal to 

-rr Kac + Kaf /q,f\ 

v a . = Kab —^ (<*4) 

In addition, the sum of all column shears in a story equals the total wind 
shear. From this requirement together with equation (34) , shears and sub- 
sequently moments may be calculated in the columns. Shears and moments 
may then be computed in the beams by using the assumption of contra- 

flexure at mid span. 

Let the dotted line extending downward past D in Fig. 30 indicate that 
AD' is a basement story column, with height h' and assumed fixed at the 
footing, D'. The regular distribution of wind pressure may be applied in 
this case also, except that for columns as AD' } the effective story height, 
h, should be taken somewhat smaller than the actual story height h', say, 
h = % X h'. 

♦Equation (34) gives shear in column AB above joint A. For the column below, AD, substitute K AD 
for K AB in equation (34). 



r ,5 



BIBLIOGRAPHY 

Vertical Load Analysis 

1. "Analysis of Statically Indeterminate Structures by the Slope-Deflection Method/' 

W. M. Wilson, F. E. Richart and Camillo Weiss, Bulletin 108, Univ. of 111., 
Eng. Exp. Sta., 1918. 

2. "Reinforced Concrete Design," Oscar Faber, published by Arnold, London, 

Vol. 2, 1924. 

3. "Statically Indeterminate Stresses," Parcel and Maney, published by John Wiley 

& Sons, 1926. 

4. "Moments in Restrained and Continuous Beams by the Method of Conjugate 

Points," L. H. Nishkian and D. B. Steinman, Trans. AJS.C.E., 1927, Vol. 90, 
pp. 1-206 (inch discussion). 

5. "Continuity as a Factor in Reinforced Concrete Design," Hardy Cross, Proc. A.CJ. 9 

1929, pp. 669-711. 

6. "Simplified Rigid Frame Design," Hardy Cross, Proc. A.C.I., Vol 26, 1930, 

pp. 170-183. 

7. "Standards of Design for Concrete," U. S. Navy Dept., Bureau of Yards and Docks, 

U. S. Government Printing Office, Washington, 1930. 

8. "The Column Analogy," Hardy Cross, Bulletin 215, Univ. of 111., Eng. Exp. Sta., 1930. 

9. "Vertical-Load Analysis of Rigid Building Frames Made Practicable," John E. 

Goldberg, Engineering News-Record, November 12, 1931, pp. 770-772. 

10. "Analysis of Continuous Frames by Distributing Fixed End Moments/' Hardy 

Cross, Tram. A.S.C.E., 1932, Vol. 96, pp. 1-156 (inch discussion). 

11. "Continuous Frames of Reinforced Concrete," Cross and Morgan, published by 

John Wilev & Sons, 1932. 

12. "The Modified Slope-Deflection Equations," L. T. Evans, Proc. A.C.L, 1932, 

pp. 109-130. 

13. "Design of Continuous Frames Having Variable Moment of Inertia,' ' Thor 

Germundsson, Civil Engineering, October, 1932, pp. 647-648. 

14. "The Design of Steel Mill Buildings and the Calculation of Stresses in Framed 

Structures," Milo S. Ketchum, published by McGraw-Hill, 1932. 

15. " Continuous-Beam Analysis by Direct Distribution," J. C. Schulze and W. A. Rose, 

Engineering News-Record, December 20, 1934, pp. 782-785. 

16. "Analysis of Continuous Structures by Traversing the Elastic Curves," Ralph W. 

Stewart, Proc. A.S.C.E., October, 1934, pp. 1125-1134. 

17. "Handbook of Rigid Frame Analysis," L. T. Evans, printed by Edwards Brothers, 

Inc., Ann Arbor, Mich., 1934. 

18. "Analysis of Continuous Frames by the Method of Restraining Stiffnesses," by 

E. B. Russell, published by Ellison and Russell, San Francisco, Calif., 1934. 

19. "A Direct Method of Moment Distribution," T. Y. Lin, Proc. A.S.C.E., December, 

1934, pp. 1451-1461. (See discussion by L. E. Grinter, March, 1935, p. 425.) 

20. "Rigid-Frame Design in the Bureau of Yards and Docks," G. A. Hunt, Engineering 

News-Record, June 27, 1935, pp. 915-917. 

56 



BIBLIOGRAPHY 

Wind Stress Analysis 

21. "Wind Stresses in the Steel Frames of Office Buildings," W. M. Wilson and G. A. 

Maney, Bulletin 80, Univ. of III., Eng. Exp. Sta., 1915. 

22. "Wind Stresses in the Steel Frames of Office Buildings," Albert Smith and W. M. 

Wilson, Jl. Western Soc. of Eng., Apr., 1915, pp. 341-390. 

23. "Column Deflection Method for Designing Lateral Bracing," John C. Van der Mey 

and Felix H. Spitzer, Engineering News-Record, January 19, 1928, pp. 106-108. 

24. "Design of Tall Building Frames to Resist Wind," Morris and Rose, Bulletin 48, 

Ohio State Univ., 1929. 

25. "Wind Stresses in Tall Buildings," R. Fleming, Engineering and Contracting, June, 

1929, pp. 250-252. 

26. "Wind Stresses in Buildings," R. Fleming, published by John Wiley & Sons, 1930. 

27. "Wind on Tall Buildings," E. E. Seelye, Engineering News-Record, March 20, 1930. 

pp. 481-483. 

28. "Wind Bracing: The Importance of Rigidity in High Towers," H. V. Spurr, published 

by McGraw-Hill, 1930. 

29. Expert Testimony on Wind Design for Tall Buildings: 

I. "A Review of Current Theories," D. C. Coyle, Engineering News-Record, 
June 4, 1931, pp. 932-934. 

II. "Assumed Loads and Fiber Stresses," Albert Smith, Engineering News- 
Record, June 11, 1931, pp. 971-974. 

III. "Chord Deflections Control Web System Design," Henry V. Spurr, Engineer- 

ing News-Record, June 18, 1931, pp. 1012-1014. 

IV. "An Analysis of Some Basic Assumptions," C. T. Morris, Engineering News- 

Record, June 25, 1931, pp. 1050-1051. 

30. "Wind Stresses on Portal Frames," W. S. Gray, Concrete and Consir. Eng., 1932, 

pp. 623-631. 

31. "Deflections and Vibrations in High Buildings," L. J. Mensch, Proc. A.C.J., Vol. 28, 

1932, pp. 387-404. 

32. "Wind-Stress Analysis and Moment Distribution," R. Fleming, Engineering News- 

Record, February 9, 1933, pp. 194-195. 

33. "Wind-Bracing Problems," A. Smith, Jl. Western Soc. of Eng., February, 1933, 

pp. 1-18. 

34. "Wind Stresses by Slope Deflection and Converging Approximations," J. E. 

Goldberg, Trans. A.S.C.E., 1934, pp. 962-985. 

35. "Wind Stress Analysis Simplified," L. E. Grinter, Trans. A.S.C.E., 1934, pp. 610-669. 

Discussion closed in Proc. A.S.C.E., January, 1934, pp. 93-102. 

36. "Basis of Design for Hurricane Exposure," Albert Smith, Proc. A.C.I. , Vol. 27, 

1931, pp. 903-924. 

37. "Structural Frameworks," Thomas F. Hickerson, The University of North Carolina 

Press, 1934. 

57 



NOTES 



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